Passport to advanced mathematics

9 posts

Linear and quadratic systems of equations

Linear and quadratic systems of equations on the SAT test

SAT Subscore: Passport to Advanced Mathematics

Linear and quadratic systems of equations include 2 equations: a linear equation and a quadratic equation:
A linear equation y=mx+b is an algebraic equation in which each term has an exponent of one. It is called linear because it can be graphed as a straight line in the xy plane.
A quadratic equation y=ax2+bx+c is an equation that has a squared term (a variable multiplied by itself) and is graphed as a parabola in the xy plane. Linear and quadratic system can be solved algebraically or graphically.

Graphical solution of linear and quadratic systems of equations is at the intersection point of the line and the parabola. This intersection can happen 0, 1 or 2 times.

Algebraic solution of linear and quadratic systems of equations can be done by substituting an expression for a variable (plugging an expression instead of a variable). This will leave us with one equation with one variable that we can solve.

Before learning this topic learn linear equations topic, quadratic equations topic, graphing quadratic functions topic and systems of linear equations topic.  

Graphical solution of linear and quadratic systems of equations

To solve a system of equations graphically, graph both equations in the same coordinate system. The solutions to the system are in the points where the two graphs intersect.

Graphical solution of linear and quadratic systems of equations steps

Step 1: Graph the linear equation as a straight line in the xy plane: 

To draw the graph, mark 2 points and draw a line between them. To make sure you have found the right points it is best to mark a third point. If there is no mistake all the points will be on the line. Which points to choose? Mark the points of intersection with the x-axis (y=0) and intersection with the y-axis (x=0) and then select a third point that is located between these two points. 

For more details about graphing a line learn the graphic presentation of a linear function topic.

Step 2: Graph the quadratic equation as a parabola in the xy plane:

When we graph a quadratic function, we need to cover the vertex, the decreasing part and the increasing part. Plug some x values into the function to calculate their corresponding y values, plot the (x,y) coordinates in the xy plan and sketch a parabola through all the points.

For more details about graphing a parabola learn the graphing quadratic functions topic.

Step 3: See the coordinates of the intersection points of the line and the parabola:

The intersection can happen 0, 1 or 2 times, therefore a linear and quadratic system of equations can have 0, 1 or 2 solutions.

The graphs below show these 3 different possibilities of the solutions.

Graphical solution of a Linear and quadratic equations systems

The position of the linear equation relative to the vertex of the parabola

Given the linear equation in a form of y=b (with no slope), the number of the solutions can be determined using the vertex of the parabola.

For more details about graphing the vertex of a parabola learn the vertex point of a parabola topic.

The different locations possibilities and the resulting numbers of solution are listed below.

If the vertex point of a parabola (minimum or maximum) is equal to b (b=y of the vertex), then there is 1 intersection point between the parabola and the line (at the vertex), therefore there is 1 solution to the equation system.

Consider the following example:

How many intersection points has the system of equations y=(x+1)2-4 and y=-4?

The vertex point of the parabola is (-1,-4), there is only one vertex point on the parabola. The equation of the line is y=-4, therefore the function will intersect the parabola at the vertex point and there is 1 solution to the equations system.

See a graphical presentation after the next paragraphs. 

If the parabola has a minimum point and b is below the vertex (b<y of the vertex) there are no intersection points between the parabola and the line, therefore there are no solutions to the systems of the equation.

Consider the following example:

How many intersection points has the system of equations y=(x+1)2-4 and y=-7?

The vertex point of the parabola is (-1,-4), this is a minimum point since any value of x different from x=-1 will give us y value that is bigger then -4, therefore the parabola is located on and above x=-1. The equation of the line is y=-7, therefore the function will never intersect the parabola and there is no solution to the equations system.

If the parabola has a minimum point and b is above the vertex (b>y of the vertex) there are 2 intersection points between the parabola and the line, therefore there are 2 solutions to the systems of the equation.

Consider the following example:

How many intersection points has the system of equations y=(x+1)2-4 and y=7?

The vertex point of the parabola is (-1,-4), this is a minimum point since any value of x different from x=-1 will give us y value that is bigger then -4, therefore the parabola is located on and above x=-1. The equation of the line is y=7, therefore the function will intersect the parabola at 2 points and there are 2 solutions to the equations system.

The graphs below shows the parabola and the different lines that were introduced in the example. The intersection points between the parabola and each one of the lines are marked on the graphs.  

The position of the linear equation relative to the minimum vertex of the parabola

If the parabola has a maximum point and b is above the vertex (b>y of the vertex) there are no intersection points between the parabola and the line, therefore there are no solutions to the systems of the equation.

Consider the following example:

How many intersection points has the system of equations y=-(x-1)2+21 and y=30?

The vertex point of the parabola is (1,21), this is a maximum point since any value of x different from x=1 will give us y value that is smaller then 21, therefore the parabola is located on and below x=21. The equation of the line is y=30, therefore the function will never intersect the parabola and there is no solution to the equations system.

If the parabola has a maximum point and b is below the vertex (b<y of the vertex) there are 2 intersection points between the parabola and the line, therefore there are 2 solutions to the systems of the equation.

Consider the following example:

How many intersection points has the system of equations y=-(x-1)2+21 and y=10?

The vertex point of the parabola is (1,21), this is a maximum point since any value of x different from x=1 will give us y value that is smaller then 21, therefore the parabola is located on and below x=21. The equation of the line is y=10, therefore the function will intersect the parabola at 2 points and there are 2 solutions to the equations system.

The graphs below shows the parabola and the different lines that were introduced in the example. The intersection points between the parabola and each one of the lines are marked on the graphs.  

The position of the linear equation relative to the maximum vertex of the parabola

Algebraical solution of linear and quadratic systems of equations

Algebraic solution of linear and quadratic systems of equations can be done by substitution method. In this method we substitute an expression containing x for y (plugging an expression instead of y), so we are left with one equation containing one variable x.

Solving a system of equations by substitution steps

Step 1: Isolate y in one of the equations so you have an equation containing y=expression with x. Note that it is the easiest to isolate y because we have x2 in one of the equations and isolating x will cause opening brackets of more complex equation like (y-a)2 in the next step.

Step 2: Substitute the expression from step 1 for a variable into the other equation. Now you stay with one equation that has only one variable that can be solved.

Step 3: Solve the equation for the remaining value and plug it into one of the equations to calculate the other variable.

For a detailed explanation about the ways of solving a quadratic equation learn the solving a quadratic equation topic.

Step 4: You can check your answer by plugging its x value into the equation you didn’t use in step 3 and making sure you are getting the same y value as in step 3.

Consider the following example:  

Solve the system of equations y=3x2+5x+3 and x=y-10.

Step 1: Isolating y in the second equation:
x=y-10
y=x+10

Step 2: Plugging y=x+10 from the second equation as the substitute for y in the first equation getting
x+10=3x2+5x+3

Step 3: Solving the equation for x and then calculating y:
3x2+4x-7=0
x1=[-4+√(42-4*3*-7)]/(2*6)=(-4+√100)/6=(-4+10)/6=6/6=1
y1= x1+10=1+10=11 the first solution is (1,10)
x2=[-4√(42-4*3*-7)]/(2*6)=(-4-√100)/6=(-4-10)/6=-14/6=-7/3=-21/3
y2= x2+10=-21/3 +10=72/3 the second solution is (-21/3, 72/3)

Step 4: Checking the solutions: To calculate the values of y we plugged the values of x1 and x2 into the second equation. Now we can plug x1 and x2 into the first equation and check if we get the same values of y:
y=3x2+5x+3
y1=3*12+5*1+3=3+5+3=11
y2=3*(-7/3)2+5*-7/3+3=(3*-7*-7)/(3*3)-35/3+3=+49/3-35-3+9/3=(+49-35+9)/3=23/3=7+2/3=72/3

Determining if a given solution is a solution for a system of equations

Each answer for a system of equations must satisfy both equations, therefore me must plug the given solution into each one of the equations. Plugging the solution into just one equation is not enough.

Graphically speaking, each point that is located on the graph of the first equation satisfies the first equation and each point that is located on the graph of the second equation satisfies the second equation, but only the points that are located on the intersection points of the two equations are the answers to the system of the equations.

Consider the following example:

Are the points (5,6) and (1,2) the solutions to the following system of equations: y=x+1 y=x2-2x+3

The point (5,6):

Plugging the point (5,6) into the first equation:
y=x+1
6=5+1
6=6 is a true statement therefore we know that the solution (5,6) satisfies the first equation.

Plugging the point (5,6) into the second equation:
y=x2-2x+3
6=5*5-2*5+3
6=25-10+3
6=18 is a false statement therefore we know that the solution (5,6) doesn’t satisfy the second equation.

The answer is that the point (5,6) is not a solution for a system of the equations.

The point (1,2):

Plugging the point (1,2) into the first equation:
y=x+1
2=1+1
2=2 is a true statement therefore we know that the solution (1,2) satisfies the first equation.

Plugging the point (1,2) into the second equation:
y=x2-2x+3
2=1*1-2*1+3
2=1-2+3
2=2 is a true statement therefore we know that the solution (1,2) satisfies the second equation.

The answer is that the point (1,2) is a solution for a system of the equations.

Radical and rational equations and expressions

Radical and rational equations and expressions on the SAT

SAT Subscore: Passport to Advanced Mathematics

Solving radicals and rational exponents requires creating equations that are different from the original equations. This may result in getting extraneous solutions, meaning that the solutions to the new equations don’t satisfy the original equations and therefore are not correct.

Rational expressions: 
A rational expression is an expression containing at least one fraction with a variable in the denominator. The variables in the nominator and the denominator can be a quadratic or a higher order polynomial. For example: 3x/(x2-2) is a rational expression with linear variable in the nominator and a quadratic variable in the denominator.

Solving rational expressions includes simplifying, performing 4 operations with fractions (adding, subtracting, dividing, and multiplying) and cancelling common factors. 

Rewriting a rational expression as a quotient and a remainder can be done with 2 methods: long division and grouping the numerator. 

Rational equations: 
A rational equation is an equation containing at least one fraction with a variable in the denominator. For example: 2/(x+2)=1.

Solving rational equations is done by multiplying both sides of the equation by the least common denominator. Since rational expressions contain a variable in the denominator, we need to exclude an extraneous solution for which the denominator equals to zero (we can’t divide by 0).

Radical equations:
Radicals are rational exponents that are written with roots. For example √x. The symbol of a radical is √ and it represents a square root (instead of writing 2√x we write only √x).

A radical equation is an equation in which a variable appears under a radical. For example: √(x+1)=1 is a radical equation and √(4+12)=x is not a radical equation.

Solving radical equation is done by squaring both of its sides, this action cancels the radical sign and results in a linear or quadratic equation that we can solve. Extraneous solution is a solution that we get after solving a squared equation that is not a solution to the original equation. We need to exclude extraneous solutions since they are incorrect.

Continue reading this page for detailed explanations and examples.

Rational expressions on the SAT

A rational expression is an expression containing at least one fraction with a variable in the denominator. Rational expression can also contain variables in the nominator. The variables in the nominator and the denominator can be a quadratic or a higher order polynomial. For example:
2/(x+2) is a rational expression with a variable in the denominator (the minimal requirement for a rational expression).
2x/(x+3) is a rational expression with linear variables in the nominator and the denominator.
3x/(x2-2) is a rational expression with linear variable in the nominator and a quadratic variable in the denominator.

Simplifying a rational expression

Before performing operations with two or more rational expressions, you should check if it is possible to simplify each expression by cancelling out common factors (factors that appear in both the nominator and the denominator). Making different operations on simplified expressions will be much easier.

The steps for simplifying one rational expression:

Step 1- Check if you can factor out common factors from the nominator and the denominator. Cancel common factors that appear both in the numerator and the denominator.

Step 2- Given a quadratic expression, check if you can rewrite it to its factored form or apply special factoring. Cancel common factors that appear both in the numerator and the denominator.
For detailed explanations learn the factored form of a quadratic function subject and the special factoring subject.

Consider the following example:

Simplify the expression (x3-4x)/(x2-2x).

Step 1- Factoring out common factors:
We can factor out x in the nominator and the denominator getting (x3-4x)/(x2-2x)=x(x2-4)/x(x-2)
Cancelling common factors: x(x2-4)/x(x-2)=(x2-4)/(x-2).

Step 2- Applying special factoring:
According to the difference of squares formula a2-b2=(a+b)(a-b). We can identify this formula in the nominator of the expression (x2-4)/(x-2) since x2 is a2 and 4 is b2, therefore x2-4=(x+2)(x-2).

Cancelling common factors: (x2-4)/(x-2)=( x+2)(x-2)/(x-2)=x+2.

Solving rational expressions

After checking simplifying each rational expression as explained before, you can start performing operations with two or more rational expressions.

The steps for solving rational expressions:

Step 1- Check if you can simplify each expression:
As said before, prior to making operations with two or more rational expressions, you should first try to simplify each expression.

Step 2- Make the operation between the expressions:
Solving rational expressions questions requires performing 4 operations with fractions (adding, subtracting, dividing, and multiplying).
For detailed explanations learn the solving a linear equation with fractions topic.

Step 3- Cancel common factors that appear both in the numerator and the denominator.

Multiplying rational expressions

Consider the following example:

What is the product of (x2+2x)/(x-1) and (x3-x)/x2?

Step 1- Simplify each expression:
We can simplify the first expression by factoring out x in the nominator: (x2+2x)=x(x+2).
We can simplify the second expression by factoring out x in the nominator and then cancelling it: (x3-x)/x2=x(x2-1)/x*x=(x2-1)/x.

Step 2- Making operation between the expressions:
The multiplication between fractions formula is ab*cd=a*cb*d.
In the question [x(x+2)/(x-1)] * [(x2-1)/x] we have a=x(x+2), b=(x-1), c=(x2-1) and d=x, therefore:
[x(x+2)/(x-1)] * [(x2-1)/x] = [x(x+2)*(x2-1)] / [(x-1)*x]

Step 3- Cancelling common factors that appear both in the numerator and the denominator:
We can cancel x and x-1 getting [x(x+2)*(x2-1)] / [(x-1)*x]=(x+2)*(x-1).

Dividing rational expressions

Consider the following example:

Solve [(x2+4x+3)/(x2+4x+4] / [(x+1)/(x+2)].

Step 1- Simplify each expression:
Rewriting the expression x2+4x+3 to its factored form will give us x2+4x+3=x2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3).
Rewriting the expression x2+4x+4 according to the square of sum formula
(a+b)2=a2+b2+2ab will give us x2+4x+4=(x+2)2.
Therefore the first expression is simplified to (x+1)(x+3)/(x+2)2

Step 2- Making operation between the expressions:
We need to solve [(x+1)(x+3)/(x+2)2] / [(x+1)/(x+2)]
According to fraction division formula ab:cd=ab*dc=a*db*c, in the given question we get a=(x+1)(x+3), b=(x+2)2, c=x+1, d=x+2, therefore:
[(x+1)(x+3)/(x+2)2] / [(x+1)/(x+2)] =[(x+1)(x+3)(x+2)] / [(x+2)2(x+1)]

Step 3- Cancelling common factors that appear both in the numerator and the denominator:
We can cancel x+1 and x+2 getting (x+3)/(x+2).

Adding and subtracting rational expressions

Consider the following example:

What is the difference between (5x2-2x)/(x2+x) and 3x/(x-1)?

Step 1- Simplify each expression:
We can simplify the first expression by factoring out x in the nominator and the denominator and then cancelling it.
(5x2-2x)/(x2+x)=[x(5x-2)/x(x+1)]=(5x-2)/(x+1)

Step 2- Making operation between the expressions:
We have to solve (5x-2)/(x+1)-3x/(x-1).
First, we need to find a common denominator of the expressions using the formula for subtraction of fractions with unlike denominators ab±cd=(ad±bc)bd:
(5x-2)/(x+1)-3x/(x-1)=[(5x-2)(x-1)]/[(x+1)(x-1)][3x(x+1)]/[(x-1)(x+1)]

After finding a common denominator we can solve using the formula for addition and subtraction of fractions with like denominators ab±cb=(a±c)b :
[(5x-2)(x-1)-3x(x+1)]/[(x+1)(x-1)]=(5x2-7x+2-3x2-3x)/(x2-1)=(2x2-10x+2)/(x2-1)

We can factor out 2 in the nominator:
(2x2-10x+2)/(x2-1)=2(x2-5x+1)/(x2-1)

Rewriting a rational expression as a quotient and a remainder

A quotient is the result of dividing one number by another. If the answer is not an integer, then the quotient is only the integer part of the division. The part that is left is called a remainder.
For example:
28/7=4 the quotient of 28 and 7 is 4 (no remainder).
30/7=4+2/7 the quotient of 30 and 7 is 4 and the remainder is 2.

We can express a division of 2 polynomial expressions a and b as 2 new polynomials: a polynomial q that represents the quotient and another polynomial r  that represents the remainder:
a(x)/b(x)=q(x)+r(x)/b(x)

Note that:
a(x)>b(x), otherwise we can’t create a quotient.
q(x) has no denominator, therefore it represents the quotient; r(x) is divided by the denominator b(x), therefore r(x) is the remainder.

There are 2 methods to divide polynomials: long division and grouping the numerator. It is enough to know the long division method to solve polynomial division questions.

Long division polynomials method

When dividing 2 numbers we can relate to them as a dividend (the number that we are dividing) and a divisor (the number that divides another number). When we start to do the long division, the dividend is the numerator and the divisor is the denominator. A we continue, the dividend changes according to what is left to divide and the divisor stays equal to the denominator.

The rule in long division of polynomials is that we need to look at the highest exponents of the dividend and the divisor. For example: in the expression 5x2-2x the highest exponential term is 5x2.

long division of polynomials a(x)/b(x) steps:

Step 1: Divide the terms with the highest exponents of the dividend (the numerator a(x)) and the divisor (the denominator b(x)) and place the result above the bar.

Step 2: Multiply the result of step 1 by the denominator b(x) and place the result below the dividend adding it a minus sign (the numerator):

Step 3: Subtract the result from step 2 from the dividend a(x).

Step 4: Continue dividing replacing and subtracting the dividend with what is left from the subtraction:

Step 5: Check the answer.

Consider the following example:

Rewrite (5x3+4x2+3)/x2 as a quotient and a remainder.

Step 1: Dividing the terms with the highest exponents of the dividend (the numerator 5x3+4x2+3) and the divisor (the denominator x2) and place the result above the bar:
The term with the highest exponents of the dividend (the numerator) is 5x2.
The term with the highest exponents of the divisor (the numerator) is x2.
We need to divide 5x3/x2 and place the result 5x above the bar.

         5x
      ____________
x2  |  5x3+4x2+3

Step 2: Multiplying the result of step 1 by the denominator x2 and placing the result below the dividend (the numerator) adding it a minus sign (we add a minus because we need to subtract in the nest step):

         5x
      ____________
x2  |  5x3+4x2+3
        -5x3

Step 3: Subtracting the result from step 2 (in the previous step we added the minus sign to 5x3) from the dividend 5x3+4x2+3:

         5x
       ____________
x2  |  5x3+4x2+3
        -5x3
       _______
         4x2+3

Step 4: Continuing dividing replacing and subtracting the dividend with what is left from the subtraction:
The new dividend (instead of the numerator a(x)) is now 4x2+3, the denominator b(x) stays as the divisor b(x) which is x2. The answer is 4 and what is left is 4x2+3-4x2=3. Therefore we write -4x2 under the dividend and add 4 to the answer that is written above the bar (getting 5x+4 above the bar):

        5x+4
      ____________
x2  |  5x3+4x2+3
        -5x3
       _______
       4x2+3
      -4x2
       _______
        3

We are left with dividend of 3. We can’t divide 3 by x2, therefore the remainder is 3/ x2.
The answer is (5x3+4x2+3)/x2=5x+4+3/x2

Step 5: We can check the answer by calculating a common denominator of the answer 5x+4+3/x2:
The common denominator is x2:
(5x+4)*x2/x2+3/x2=(5x3+4x2+3)/x2 this is the original expression.

Polynomials division- factoring the numerator

In this technic we split the numerator to a multiplication of the denominator and another expression and add what is left. This will allow to cancel the denominator and get us a quotient that we need, what is left will become the remainder.

Polynomials division a(x)/b(x)- factoring the numerator steps:
Step 1: Split the numerator to a multiplication of the denominator and another expression and another expression.
Step 2: Cancel the denominator and get a quotient and a remainder.
Step 3: Check the answer.

Consider the previous example:

Rewrite (5x3+4x2+3)/x2 as a quotient and a remainder.

In this example the denominator is composed from only one component x2, therefore we can factor it out:

Step 1: Splitting the numerator to a multiplication of the denominator and another expression and another expression:
We need to split the numerator 5x3+4x2+3 x2 into the denominator x2 multiplied by another expression and some other expression. We see that we can factor out a common factor of x2 from 5x3+4x2 leaving 3, getting 5x3+4x2+3=x2(5x+4)+3.
The original expression was (5x3+4x2+3)/x2, combining that (5x3+4x2+3)/x2= [x2(5x+4)+3]/x2

Step 2: Cancelling the denominator and getting a quotient and a remainder:
[x2(5x+4)+3]/x2=5x+4+3/x2 the quotient is 5x+4 and the remainder is 3.

Step 3: We can check the answer by calculating a common denominator of the answer 5x+4+3/x2:
The common denominator is x2:
(5x+4)*x2/x2+3/x2=(5x3+4x2+3)/x2 this is the original expression.

Consider the following example:

Rewrite (5x3+4x2+3)/x2 as a quotient and a remainder.

In this example the denominator is composed from an expression x-1, therefore we need to factor out x-1 from the nominator. For detailed explanations learn the factored form of a quadratic function subject.

Step 1: Splitting the numerator to a multiplication of the denominator and another expression and another expression:

We have x2+3 and we need to split it to a multiplication of x-1 by another expression and add what is left: x2+0x+3=(x-1)(x+?)+??

We need 2 numbers that add to 0 and we know that one of them is -1. Therefore the other number is 1. We get (x-1)(x+1)=x2-x+x-1= x2-1, the full numerator was x2+3 therefore we are left with 4 that we couldn’t factor: x2+3=(x-1)(x+1)+4.

Step 2: Cancelling the denominator and getting a quotient and a remainder:
(x2+3)/(x-1)=[(x-1)(x+1)+4]/(x-1)= [(x-1)(x+1)]/(x-1)+4/(x-1)=x+1+4/(x-1) the quotient is x+1 and the remainder is 4.

Step 3: We can check the answer by calculating a common denominator of the answer x+1+4/(x-1):
The common denominator is x-1:
x+1+4/(x-1)=(x+1)(x-1)/(x-1)+4/(x-1)=[(x+1)(x-1)+4]/(x-1)=(x2-x+x-1+4)/(x-1)=(x2+3)/(x-1) this is the original expression.

 

Rational equations on the SAT

Solving rational equations steps

Step 1: Reaching a common denominator to cancel the denominators: Reaching a common denominator is done by multiplying by the smallest common multiple of the denominators of the fractions.

Step 2: Solving the equation: We can solve the new equation since it does not include denominators, it can be a linear equation or a quadratic equation.

Consider the following example: 

Solve the equation 2/(x-1)=3/x.

2/(x-1)=3/x

The common denominator is x(x-1):
2x(x-1)/(x-1)=3x(x-1)/x
2x=3(x-1)
2x=3x-3
-x=-3
x=3

Checking the answer:
2/(x-1)=3/x
2/2=3/3
1=1 is a true statement therefore the answer x=3 is correct.

Step 3: Excluding extraneous solutions:

Plug the solutions of the equation without denominators into the original equation to make sure that they satisfy the original equation.

Sometimes the answers of the equation without denominators don’t satisfy the original equation. See the next paragraph for further explanation.

Extraneous solutions of rational equations

Extraneous solution is a solution that we get after solving a squared equation that is not a solution to the original equation. We need to exclude extraneous solutions since they are incorrect. To find if the solution is extraneous check the solutions by plugging it into the original equation (the solution is extraneous if it does not satisfy the equation).

When do we get extraneous solutions? Extraneous solution happens when a solution for the equation without denominators causes a division by 0 in the original equation.  

Consider the following example:

Solve 4/(x-3)=(x+1)/(x-3).

4/(x-3)=(x+1)/(x-3)

The common denominator is (x-3):
4(x-3)/(x-3)=(x+1)(x-3)/(x-3)
4=x+1
x=3

Checking the answer:
4/(3-3)=(x+1)/(3-3) the denominator on both sides of the equation is 3=3=0, we cannot divide by 0 therefore x=3 is an extraneous solution and there is no solution to the equation.
 

Radical equations on the SAT

Solving radical equations steps

To solve the radical equation, we need to cancel the radical. This can be done by raising both sides of the equation to the index of the radical. Usually, the radical is a square root, therefore we need to raise it to a second exponent. Squaring both sides of the equation leads us to a new equation that we need to solve.

Step 1- Isolate the radical expression on one side of the equation:

If we have an expression that includes other components in addition to the radical we can’t cancel the radical in the next step.
For example: in the equation √x+1=1 the radical is not isolated in one side of the equation therefore we cannot cancel the radical by squaring both parts of the equation:
√x+1=1
Squaring both sides: (√x+1)2=12
Remember the formula (a+b)2=a2+b2+2ab therefore (√x+1)2=x+1+2√x.
The new squared equation is x+1+2√x=1, the radical sign still exists in the equation.
Isolating the radical on one side of the equation at first gives us:
√x+1=1
√x=1-1
√x=0
In this arrangement of the equation (the radical was isolated on one side of the equation before squaring), squaring both parts of the equations will cancel the radical:
√x=0
(√x)2=0
x=0

Step 2: Square both sides of the equation:

If we raise the radical of a square root to a second exponent, we get only the expression under the radical, since (√x)2= x1/2*2= x2/2=x1=x. Since there is no radical in the equation, we can solve it in the next step.

Step 3: Solve the new squared equation:

We can solve the new equation since it does not include radicals. It can be a linear equation or a quadratic equation.

We will get a linear equation if on the other side of the equation there is a constant (squaring a constant will give us another constant). For example:
√(2x+4)=4 (4 is a constant)
(√(2x+4))2=42 (42 is also a constant)
2x+4=16 is a linear equation
2x=12
x=6

We will get a squared equation if there is a squared variable under the radical. For example: √(x2-2x)=-1.
Another case in which we will get a squared equation is when on the other side of the original equation there is a variable or an expression with variables (squaring a variable will give us a squared variable). For example:
√(5x-4)=x (x is a variable)
(√(5x-4))2=x2
5x-4=x2 is a squared equation

Solving the squared equation ax2+bx+c=0 is done using the quadratic formula
x=(-b±√(b2-4ac))/2a,    x1=(-b+√(b2-4ac))/2a,    x2=(-b-√(b2-4ac))/2a
5x-4=x2
-x2+5x-4=0
x1=(-5+√(25-4*4))/-2=(-5+√9)/-2=-2/-2=1
x2=(-5-√(25-4*4))/-2=(-5-√9)/-2=-8/-2=4

Checking the answers:
For x=1 we get 5-4=1, 1=1
For x=4 we get 20-4=16, 16=16

Step 4: Excluding extraneous solutions:

Plug the solutions of the squared equation into the original equation to make sure that they satisfy the original equation.

Sometimes the answers of the squared equation don’t satisfy the original equation. See the next paragraph for further explanation.

Extraneous solutions of a radical equations

Extraneous solution is a solution that we get after solving a squared equation that is not a solution to the original equation. We need to exclude extraneous solutions since they are incorrect. To find if the solution is extraneous check the solutions by plugging it into the original equation (the solution is extraneous if it does not satisfy the equation).

Why do we get extraneous solutions? The reason is that the squared equation is different from the original equation, as a result it has more solutions. 

For example: the original equation is x=2 (the solution is 2), its squared equation is x2=4 (the solution is 2 or -2). After squaring the original equation x=2 to a new equation x2=4 we get 2 solutions 2 or -2, whole the only solution for the original equation x=2 is 2. Squaring got us another answer x=-2 that is an answer for the squared equation but not an answer for the original equation. Since our goal was to solve the original equation, we must exclude the extraneous answer x=-2.   

Consider the following example:

What are the solutions to the equation √(-3x-2)=x?

√(-3x-2)=x

Squaring both sides of the equation: (√(-3x-2))2=x2
-3x-2=x2
x2+3x+2=0
x1=(-3+√(9-4*2))/2=(-3+1)/2=-1
x2=(-3-√(9-4*2))/2=(-3-1)/2=-2

Checking the answers by plugging them into the original equation:
For x=-1 we get √(-3*-1-2)=-1, √1=-1, 1=-1 is a false statement therefore the solution x=-1 is extraneous and we need to exclude it.
For x=-2 we get √(-3*-2-2)=-2, √4=-2, 2=-2 is a false statement therefore the solution x=-1 is extraneous and we need to exclude it.
The original equation √(-3x-2)=x has no solutions while the quadratic equation had 2 solutions.

 

Polynomial functions and graphs

Polynomial functions and graphs on the SAT

SAT Subscore: Passport to Advanced Mathematics

Polynomial as a mathematical expression made up of more than one term, where each term has a form of axn (for constant a and none negative integer n). For example: 2x3.

In polynomial function the input is raised to second power or higher. The degree of a polynomial function is defined as its highest exponent.
Even degree polynomial function has an even highest exponent (2, 4, 6, etc.).
Odd degree polynomial function has an odd highest exponent greater than 1 (3, 5, 7, etc.).

For example: The function f(x)=x3+3x2-x-3 is a third degree polynomial function (in this function the highest exponent is 3), it is an odd degree function (the highest exponent 3 is odd).

The factored form of a polynomial function shows the x intercepts of the function.

The standard form of the polynomial function shows the y intercept of the function.

The end behavior shows the location of the graph of the function for very small and very large values of x. To know the end behavior of a polynomial function we need to look at its highest exponent (if it is odd or even) and the sign of its coefficient.

The polynomial remainder theorem says that when we divide a polynomial function f(x) by the expression x-a the remainder is f(a). Therefore, to find the remainder we do not need to do the division, we just plug x=a into f(x) and calculate the output.

Note that polynomial is written in descending order of its exponents.

Quadratic functions are the simplest form of polynomial functions (they have an exponent of 2). Quadratic functions topic is divided into two pages: quadratic equations and quadratic functions and graphing quadratic functions.

Another topic that you should know before learning about polynomial functions is exponential expressions topic, since it explains the rules and the operations with exponents.

Continue reading this page for detailed explanations and examples.

The factored form of a polynomial function and its x intercepts

The factored form of a third degree polynomial function is f(x)=(x-a)(x-b)(x-c). For example: f(x)=(x+1)(x-1)(x+3) is the factored form of a standard form function f(x)=x3+3x2-x-3. Note that we can open brackets of the factored form and get the standard form.

X intercepts are the points where the function crosses the x axis, x intercepts are also called zeros and roots. The maximum number of x intercepts of a polynomial function is equal to the degree of the function. The minimum number of x intercepts is zero for an even degree polynomial function and 1 for an odd degree polynomial function.

A zero product property states that for any degree polynomial function, the number of x intercepts in equal to the number of the factors in the function. For example: For a third degree polynomial function with a factored form f(x)=(x-a)(x-b)(x-c) a zero product property states that if a multiplication of 3 numbers is equal to 0 then one of the numbers must be equal to 0. In a factored equation (x-a)(x-b)(x-c)=0 that means that x-a=0 or x-b=0 or x-c=0. Since a, b and c are constants that we know, we can see the 3 x intercepts values from the factored formula, so that the x intercepts are (a,0), (b,0) and (c,0).

To find the x intercepts given the factored form of a polynomial function, we need to set each factor to be equal to zero and then solve.

Consider the following example:

What are the x intercepts of the function f(x)=(x+1)(x-1)(x+3)?

x+1=0, x=-1
x-1=0, x=1
x+3=0, x=-3

The x intercepts of the function are at the points (-1,0), (1,0) and (-3,0).

The graph below represents the function f(x)=(x+1)(x-1)(x+3).
The table near the graph shows different (x,y) values that were used to plot the graph.
The x axis intercepts are marked in red on the graph and in the table.

polynomial function- factored form and x intercepts

Writing the factored form of a polynomial function from its intercepts:

For any intercept value a, the corresponding expression in the factored function will be x-a, so that x-a=0. For example: if the x intercept is x=4 then the factored expression is x-4, since 4-4=0.

In an polynomial function of a third degree we are given the x intercepts values as a, b and c, therefore the factored form is f(x)=(x-a)(x-b)(x-c).

Consider the following example:

The x intercepts of a polynomial function are x=-3, x=2 and x=-1.

What it the standard form of the function?

First we will find the factored form and then open brackets to find the standard form.
For x=-3 the factored expression is x+3 since -3+3=0.
For x=2 the factored expression is x-2 since 2-2=0.
For x=-1 the factored expression is x+1 since -1+1=0.
The factored function is f(x)=(x+3)(x-2)(x+1).

Finding the standard form from the factored form:
Opening brackets will give us
f(x)=(x+3)(x-2)(x+1)
f(x)=(x2-2x+3x-6)(x+1)
f(x)= x3-2x2+3x2-6x+x2-2x+3x-6

The standard form is f(x)= x3+2x2-5x-6.

The standard form of a polynomial function and its y intercept

The standard form of a polynomial function of a third degree is f(x)=ax3+bx2+cx+d.

The y intercept coordinate is f(x=0)=d, therefore the y intercept of a polynomial function is its constant term d.

The end behavior of a polynomial function

The end behavior shows the location of the graph of the function for very small and very large values of x.

To know the end behavior of a polynomial function we need to look at its highest exponent (if it is odd or even) and the sign of its coefficient. The reason that we look only at the highest exponent is that when dealing with large positive or negative numbers the highest exponent dominates the output value of the function. For example, given x=10 we get x3=103=1,000 and x2=102=100.

For example: In the standard form of a third degree polynomial function f(x)=ax3+bx2+cx+d the highest exponent is 3 and its coefficient is a. We know that 3 is odd and that a can be positive or negative.

The end behavior of an even polynomial function

If the function is even (its highest exponent is 2, 4, 6, etc.), the ends of the graph approach the same direction:

The form of a fourth degree polynomial function is f(x)=ax4+bx3+cx2+dx+e.

If a>0 both ends of the function approach a positive infinity. The reason is that we have a positive number a>0 multiplied by a positive number (any number x raised to an even exponent will get a positive output).

For example (a>0 and x>0):
Given the function f(x)=x4+x3+2x2-5x-6 for x=10 we get
f(x=10)=104+103+2*102-5*10-6=10,000+1,000+200-50-6=11,144 (the dominant expression is x4=10,000)

For example (a>0 and x<0):
Given the function f(x)=x4+x3+2x2-5x-6 for x=-10 we get
f(x=-10)=-104-103+2*-102-5*-10-6=10,000-1,000+200+50-6=9,244 (the dominant expression is x4=10,000)

If a<0 both ends of the function approach a negative infinity. The reason is that we have a negative number a<0 multiplied by a positive number (any number x raised to an even exponent will get a positive output).

For example (a<0 and x>0):
Given the function f(x)=-x4+x3+2x2-5x-6 for x=10 we get
f(x=10)=-104+103+2*102-5*10-6=-10,000+1,000+200-50-6=-8,856 (the dominant expression is -x4=-10,000)

For example (a<0 and x<0):
Given the function f(x)=-x4+x3+2x2-5x-6 for x=-10 we get
f(x=-10)=-1*-104-103+2*-102-5*-10-6=-10,000-1,000+200+50-6=-10,756 (the dominant expression is -x4=-10,000)

The graph below represents the functions f(x)=-x4+x3+2x2-5x-6 and f(x)=x4+x3+2x2-5x-6. The table near the graph shows different (x,y) values that were used to plot the graph. The points that were explained in the above examples are marked in red in the table.

The end behavior of the graph of an even polynomial function

The end behavior of an odd polynomial function

If the function is odd (its highest exponent is 3, 5, 7, etc.), the ends of the graph approach different directions:

If a>0:

Y approaches infinity as x increases. The reason is that we have a positive number a>0 multiplied by a positive number (since x is positive).
For example: given the function f(x)=x3+2x2-5x-6 for x=10 we get
f(x=10)=103+2*102-5*10-6=1,000+200-50-6=1,144 (the dominant expression is x3=1,000)

Y approaches a negative infinity as x decreases. The reason is that we have a positive number a>0 multiplied by a negative number (since x is negative and the exponent is odd).
For example: given the function f(x)=x3+2x2-5x-6 for x=-10 we get
f(x=-10)=-103+2*-102-5*-10-6=-1,000+200+50-6=-756 (the dominant expression is -x3=-1,000)

If a<0:

Y approaches infinity as x decreases. The reason is that we have a negative number a<0 multiplied by a negative number (since x is negative and the exponent is odd).
For example: given the function f(x)=-x3+2x2-5x-6 for x=-10 we get
f(x=-10)=-1*-103+2*-102-5*-10-6=1,000+200+50-6=1,244 (the dominant expression is -x3=1,000)

Y approaches a negative infinity as x increases. The reason is that we have a negative number a<0 multiplied by a positive number (since x is positive).
For example: given the function f(x)=-x3+2x2-5x-6 for x=10 we get
f(x=-10)=-103+2*102-5*10-6=-1,000+200-50-6=-856 (the dominant expression is -x3=-1,000)

The graph below represents the functions f(x)=-x3+2x2-5x-6 and f(x)=x3+2x2-5x-6. The table near the graph shows different (x,y) values that were used to plot the graph. The points that were explained in the above examples are marked in red in the table.

The end behavior- an odd polynomial function

The polynomial remainder theorem

A remainder is what is left after dividing one number by another. For example: if we divide 5 by 2 the remainder is 1 (5/2=4/2+1).

The polynomial remainder theorem says that when we divide a polynomial function f(x) by the expression x-a the remainder is f(a). Therefore, to find the remainder we do not need to do the division, we just plug x=a into f(x) and calculate the output.

Note that:

  • To use the theorem we need to divide only by an expression in a form of x-a. For example: if dividing by x+1 the parameter a=-1, if dividing by x-5 the parameter a=5.
  • The theorem deals only with the remainder and does not tell us what is the factor of x-a.

For example: in the division of the function f(x)=(x+1)(x+1)(x-4)+3 by x-4 we see a multiplication of x-4 by the function (x+1)(x+1) and a remainder which is 3 (the number 3 is not multiplied by x-4 therefore it is the remainder).

In order to test the theorem with different examples we will use the function f(x)=(x+1)(x+1)(x-4)+5. Since the function is already written in the form of a multiplication of x-4 and x+1 by another function, we know that if we divide by x+1 or by x-4 then the remainder is 5. To test the theorem, we will first write the factored function in its standard form:
f(x)=(x+1)(x+1)(x-4)+5
f(x)=(x2+x+x+1)(x-4)+5
f(x)=x(x2+2x+1)-4(x2+2x+1)+5
f(x)=x3+2x2+x-4x2-8x-4+5

f(x)=x3-2x2-7x+1 is the standard form of the function f(x)=(x+1)(x+1)(x-4)+5.

Consider the following example:

What is the remainder of f(x)=x3-2x2-7x+1 divided by x-4?

In this example x-c=x-4 therefore c=4 since x-c=x-4.

According to the polynomial remainder theorem when we divide the function f(x)=x3-2x2-7x+1 by x-4 the remainder is f(c) which is f(4).

f(4)=43-2*42-7*4+1=64-2*16-28+1=64-32-28+1=5

The remainder according to the theorem is 5, as we saw in the form f(x)=(x+1)(x+1)(x-4)+5. The theorem helps us to find the remainder 5 and does not deal with finding the factor of x-4  which is (x+1)(x+1).

Consider the following example:

What is the remainder of f(x)=x3-2x2-7x+1 divided by x+1?

In this example x-c=x+1 therefore c=-1 since x-c=x–1=x+1.

According to the polynomial remainder theorem when we divide the function f(x)=x3-2x2-7x+1 by x+1 the remainder is f(c) which is f(-1).

f(4)=-13-2*-12-7*-1+1=-1-2+7+1=5

The remainder is 5, as we saw in the form f(x)=(x+1)(x+1)(x-4)+5. The theorem helps us to find the remainder 5 and does not deal with finding the factor of x+1 which is (x-4)(x+1)).  

A division with a remainder of zero, f(a)=0

If the remainder f(a)=0 then the function has a factored form of the expression x-a multiplied by another function q(x), meaning that x-a is a factor of f(x):
f(x)=(x-a)*q(a)+0.

To find the x intercepts we need to solve f(x)=(x-a)*q(x)=0. According to the zero product property, it can happen if q(x)=0 or x-a=0. Therefore the point x=a is an x intercept (x-a=a-a=0).

Graphing exponential functions

Graphing exponential functions on the SAT

SAT Subscore: Passport to Advanced Mathematics

Exponential function is a function with a positive constant other than 1 raised to an exponent that includes a variable.

The basic form of the exponential function is f(x)=bx (b is the base and x is the exponent).
The base b is always positive (b>0) and not equal to one (b≠1).
For example: the function f(x)=3x is an exponential function where the base is a constant b=3 and the exponent is the variable x.

The y axis intercept of the basic exponential function graph f(x)=bx is equal to 1 for all values of b.

An exponential function slope is always increasing or always decreasing. The slope form depends on the value of the base b. All graphs of exponential functions are curved.

The ends of the exponential function graph: The graph of a basic exponential function f(x)=bx has a horizontal asymptote on one of its ends (positive x axis or negative x axis). The other end of the function approaches infinity. The end behavior depends on the value of the base b.

We can shift an exponential function graph by adding a constant to the function or by multiplying the exponential term by a coefficient, so that the basic function f(x)=bx will become f(x)=a*bx+d.

Continue reading this page for detailed explanations and examples.

Two types of exponential functions

The form of the graph of the basic exponential function f(x)=bx depends on the size of the base b:

If b>1 then the slope of the graph is always positive (the graph is increasing), as x increases the graph approaches infinity and as x decreases the graph approaches zero (horizontal asymptote at x=0).

 If 0<b<1 then the slope of the graph is always negative (the graph is decreasing), as x increases the graph approaches zero (horizontal asymptote at x=0) and as x decreases the graph approaches infinity.

Continue reading this page for detailed explanations and examples.

Exponential functions- different values of the base

The features of exponential functions

Y intercept of an exponential function

Y intercept this is the point where the function crosses the y axis.

To find the y intercept, we set x=0 and evaluate the function.

Note that any number raised to a power of 0 is equal to 1, therefore in the basic function f(x)=bx the intercept is equal to 1 for all values of b.

Consider the following example:

Find the y intercepts of a functions f(x)=3x+1 and f(x)=3x-5.

f(x)=3x+1
f(0)=30+1=1+1=2

f(x)=3x-5
f(0)=3x-5=1-5=-4

The slope of an exponential function

An exponential function slope is always increasing or always decreasing, therefore, to find the slope of the function we only need to evaluate 2 points from the function.

Let’s mark point 1 (x1,y1) and point 2 (x2, y2).

If the y value increases when the x value increases (exponential grows), then the slope of the function is positive, and the function graph is always increasing (meaning that x2>x1 and y2>y1).

If the y value decreases when the x value increases (exponential decay), then the slope of the function is negative, and the function graph is always decreasing (meaning that x2>x1 and y2<y1).

Consider the following example:

Determine if the slope of the function f(x)=3x+1 is negative or positive. 

f(x)=3x+1
f(0)=30+1=1+1=2
f(1)=31+1=3+1=4

The points are (0,2) and (1,4). We got a bigger y for a bigger x, therefore the slope of the function is positive.

Consider the following example:

Determine if the slope of the function f(x)=0.5x+1 is negative or positive. 

f(x)=0.5x+1
f(0)=0.50+1=1+1=2
f(1)=0.51+1=1.5

The points are (0,2) and (1,1.5). We got a smaller y for a bigger x, therefore the slope of the function is negative.

The graphs below show the functions f(x)=0.5x+1 and f(x)=3x+1.

The points x+0 and x=1 that were used to find the slopes are marked in red on the graphs.

The slope of an exponential functions graphs

The horizontal asymptote- the end behavior

Asymptote is a line that the graph approaches but never touches.

Horizontal asymptote is a horizontal line that the graph approaches when x gets very large or very small.

The graph of a basic exponential function f(x)=bx has a horizontal asymptote on one of its ends (positive x axis or negative x axis):

If b>1 then the function f(x)=bx has a horizontal asymptote at the negative end of the x axis.
When we raise a base that is bigger than 1 to a high negative exponent the output is a very small number, therefore the output of the function f(x)=bx is very close to zero.
For example:
Given the function f(x)=bx
if x=-4 and b=3 we get f(x)=3-4=1/34=0.012
if x=-10 and b=3 we get f(x)=3-10=1/310=0.00002

If 0<b<1 then the function f(x)=bx has a horizontal asymptote at the positive end of the x axis.
When we raise a base that is smaller than 1 to a high positive exponent the output is a very small number, therefore the output of the function f(x)=bx is very close to zero.
For example:
Given the function f(x)=bx
if x=4 and b=0.5 we get f(x)=0.54=0.5*0.5*0.5*0.5=0.063
if x=10 and b=0.5 we get f(x)=0.510=0.5*0.5*0.5*0.5…*0.5=0.00098

Consider the following example:

Determine the asymptotes of the functions f(x)=3x and f(x)=0.5x.

The graphs below show the functions f(x)=0.5x and f(x)=3x.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The horizontal asymptote x=0 is marked in red on the graph and inside the table.

Horizontal asymptote of an exponential function graphs

For the graph f(x)=3x small x values will get an output that is close to zero, therefore the function has a horizontal asymptote at x=0.

For the graph f(x)=0.5x big x values will get an output that is close to zero, therefore the function has a horizontal asymptote at x=0.

Horizontal asymptote of a function with a constant term f(x)=bx+d

In this type of functions, a constant term d is added to the basic function f(x)=bx so we get a function form of f(x)=bx+d.

When the function has a constant term, the asymptote will approach the constant instead of zero.

For example: b values of 0<b<1:
if x=4, b=0.5 and c=5 we get f(x)=0.54+5=5.063
if x=10, b=0.5 and c=-4 we get f(x)=0.510-4=-4.00098

For example: b values of b>1:
if x=-4, b=3 and c=5 we get f(x)=3-4+5=1/34+5=5.012
if x=-10, b=3 and c=-4 we get f(x)=3-10-4=1/310-4=-4.00002

 Consider the following example:

Determine the asymptotes of the functions f(x)=3x+1 and f(x)=0.5x+1.

The graphs below show the functions f(x)=0.5x+1 and f(x)=3x+1.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.

The horizontal asymptote x=1 is marked in red on the graph and inside the table.

Horizontal asymptote of an exponential function with a constant graph

For the graph f(x)=3x+1 small x values will get an output that is close to zero plus 1, therefore the function has a horizontal asymptote at x=1.

For the graph f(x)=0.5x+1 big x values will get an output that is close to zero plus 1, therefore the function has a horizontal asymptote at x=1.

Graphing an exponential function steps

An exponential function f(x)=bx behavior is divided into 2 areas by the y axis: a positive x area and a negative x area. Therefore, to graph an exponential function, we need to include a point from each area.

The steps for graphing an exponential function:
Step 1: Plotting points:
Evaluate and plot 3 points- its y intercept, a point with a positive x value (like x=1) and a point with a negative x value (like x=-1).
Step 2: Sketching a curve:
Sketch a curve between the 3 points and extend it on both sides. One end will approach a horizontal asymptote of zero along the x axis (if the graph is f(x)=bx+d the asymptote will be x=k instead of x=0). The other and will approach infinity along the y axis.

The graphs below show the functions f(x)=0.5x and f(x)=3x.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The 3 points from the different areas that were needed to plot the graphs (x=0, x=3 and x=-3) are marked in red and blue in the table and on the graphs.

exponential function graphing steps

Both functions f(x)=3x and f(x)=0.5x have a y intercept at a point (1,0).

The green function f(x)=3x where b>1 has big y outputs at positive x values (approaching to infinity) and small y outputs at negative x values (approaching to zero).

The orange function f(x)=0.5x where 0<b<1 has big y outputs at negative x values (approaching to infinity) and small y outputs at positive x values (approaching to zero).

Shifting an exponential function

We can shift an exponential function by adding a constant to the function or by multiplying the exponential term by a coefficient.  

Shifting an exponential function- adding a constant term

To shift an exponential function up we need to add a constant term to the basic function f(x)=bx getting f(x)=bx+d.

To shift an exponential function down we need to subtract a constant term from the basic function f(x)=bx getting f(x)=bx-d.

The shifting results:
The y axis intercept shifts up or down d units.
The asymptote shifts up or down d units.
The other end of the function that approaches infinity remains unchanged.

In the basic function f(x)=bx the horizontal asymptote is x=0 and the y intercept is y=1.
In the function f(x)=bx+d the horizontal asymptote will be x=d and the y intercept will be y=1+d.
In the function f(x)=bx-d the horizontal asymptote will be x=-d and the y intercept will be y=1-d.

The graphs below show the functions f(x)=0.5x , f(x)=0.5x+10, f(x)=3x and f(x)=3x+10.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.

Shifting an exponential function adding a constant to the function

The y axis intersection points are marked in red in the table and on the graph. The y axis intersection point before shifting was y=1 and after the shifting it became y=d+1=10+1=11.

Every point on both functions shifted up by the size of the constant d (d=10), you can see the change comparing the columns of the tables. Two points from each function are marked in blue in the tables and on the graphs as an example. 

The asymptotes of both functions shifted up by the size of the constant d (d=10), they are marked in red in the table and on the graphs.

Shifting an exponential function- multiplying by a coefficient

We can shift an exponential function by multiplying the exponential term by a coefficient so that the function f(x)=bx will become f(x)=a*bx.

The shifting results:
The y axis intercept will become a*1=a.
The asymptote remains unchanged (x=0).
The other end of the function that approaches infinity remains unchanged.

The graphs below show the functions f(x)=0.5x , f(x)=5*0.5x, f(x)=3x and f(x)=5*3x.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.

Shifting an exponential function- multiplying the function by a coefficient

The y axis intersection points are marked in red in the table and on the graph. The y axis intersection point before shifting was y=1 and after the shifting it became y=a*1=5.

Note that the shifting size of each point is different, since bigger x values result in a bigger output and therefore a bigger shifting, you can see the change comparing the columns of the tables.

The asymptotes of both functions remained unchanged, they are marked in black in the tables and on the graphs.

Shifting an exponential function- adding a constant term and multiplying by a coefficient

The basic function f(x)=bx will become f(x)=a*bx+d.

The shifting results:
The y axis intercept will become a*1+d=a+d.
The asymptote will be x=d.
The other end of the function that approaches infinity remains unchanged.

The graphs below show the functions f(x)=0.5x , f(x)=5*0.5x+10, f(x)=3x and f(x)=5*3x+10.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.

Shifting an exponential function- adding a constant term and multiplying

The y axis intersection points are marked in red in the table and on the graphs. The y axis intersection point before shifting was y=1 and after the shifting it became y=a*1+d=5*1+10=15.

Note that the shifting size of each point is different, since bigger x values result in a bigger output and therefore a bigger shifting, you can see the change comparing the columns of the tables.

The asymptotes of both functions shifted up by the size of the constant d (d=10), they are marked in black in the table and on the graphs.

Graphing quadratic functions

Graphing quadratic functions on the SAT

SAT Subscore: Passport to Advanced Mathematics

Graphing quadratic functions topic includes two parts

Part 1- The features of the graphs of the quadratic functions– x intercepts, y intercept, vertex points, vertical symmetry and width.

Part 2- Three types of transformations of the parabola in the xy plane:
Parabola translation– Shifting the graph up, down, to the left or to the right without changing its width, so that the distance that each point moves is the same.
Parabola translation is done by adding or subtracting  a constant c to the function or from the x variable

Parabola stretching– Changing the graph’s width leaving the x intercepts coordinates and the axis of symmetry of the parabolas the same (the distance that each point moves is not the same).
Parabola stretching is done by multiplying the function by a constant:

Parabola reflecting– Reflecting the graph across the x axis or across the y axis.
Parabola reflecting is done by multiplying the function or the x variable by -1.

Graphing quadratic functions versus linear functions

Graphing a quadratic function is different from graphing a linear function.

As we learned the graph of a linear function y=mx+b is graphed as a straight line. The x2 term that is added in the quadratic function changes the graph to a form of a U shaped symmetrical curve. This graph of the quadratic function is called a parabola and has a vertex (a minimum or a maximum) and a y intercept.

The graph below shows the graph of a quadratic function y=x²+2x+10 and a linear function y=2x+10.

parabola and linear line graphs

The difference between the functions is only the x² term. The table near the graphs shows different (x,y) values that were taken to plot the graphs.  The only common point between the two graphs is when x=0, since then the term x² is equal to zero and does not affect the position of the parabola. This point (0,10) is also the y axis intercept.

The steps for graphing quadratic functions:

When we want to graph a linear function, we need to draw a straight line between only 2 points. When we graph a quadratic function, we need more points since we need to cover the vertex, the decreasing part and the increasing part.

The steps for graphing a quadratic function are:
Step 1: Plug some x values into the function to calculate their corresponding y values.
Step 2: Plot the (x,y) coordinates in the xy plane.
Step 3: Sketch a parabola through all the points.

The features of the graphs of quadratic functions:

Y intercept of a parabola

  • Y intercept this is the point where the parabola crosses the y axis.
  • Each parabola has a y intercept.
  • There is only one y intercept for each parabola.
  • In order to find the y intercept given the standard form of a quadratic function, we set x=0 and solve for y.

Consider the following example:

Graph the y intercept of a function y=x²+2x+10.

In order to find the y intercept, we set x=0 and solve for y:
y= x²+2x+10
y= 0²+2*0+10=10

The graph below shows the function y=x²+2x+10.

X intercept of a parabola

  • X intercept this is the point where the parabola crosses the x axis.
  • A parabola can have 0-2 x intercepts.
  • X intercept is also called zero and root.

We can see the x intercepts from the factored form of a quadratic function. Given the factored form of y=a(x-b)(x-c), the intercepts are (b,0) and (c,0). For more details learn about finding x intercepts from a factored form of a quadratic equation topic.

In order to find the x intercepts given the standard form of a quadratic equation y= ax2+bx+c, we set y=0 and solve the quadratic equation ax2+bx+c=0 for x:

  • If the equation has no real solution, then the parabola has no x intercepts.
  • If the equation has 1 real solution, then the parabola has 1 x intercepts.
  • If the equation has 2 real solutions, then the parabola has 2 x intercepts.

For more details about solving a quadratic equation learn the finding x intercepts from a factored form of a quadratic equation topic.

The graphs below show 3 functions with a positive x2 term (a>0):

y=x²+2x+1, y=2x²+x+1 and y=x²-2x-8.

The table near the graphs shows different (x,y) values that were taken to plot the graphs.

The x intercept points are marked in red inside the table and on the graphs.

x intercepts of a convex parabola

The equation y=2x²+x+1=0 has no solution, you can see that its graph y=2x²+x+1 is located above the x axis, therefore it has no x intercepts.

 The equation y=x²+2x+1=0 has 1 solution, you can see that its graph y=x²+2x+1 intersects the x axis at 1 point (-1,0).

The equation y=x²-2x-8=0 has 2 solutions, you can see that its graph y= x²-2x-8 intersects the x axis at 2 points (-4,0) and (2,0).

See solving of these equations above in solving a quadratic equation topic.

The graphs below show 3 functions with a negative x2 term (a<0):
y=-x²+2x-5, y=-x²+2x-1 and y=-x²+2x+8.

The table near the graphs shows different (x,y) values that were taken to plot the graphs.

The x intercept points are marked in red inside the table and on the graphs.

x intercepts of a parabola -3 concave parabolas

The equation y=-x²+2x-5 has no x intercepts.
The equation y=-x²+2x-1 intersects the x axis at 1 point (1,0).
The equation y=-x²+2x+8 intersects the x axis at 2 points (-2,0) and (4,0).

A vertex point of a parabola

  • The vertex is the lowest point (minimum point) or the highest point (maximum point) of the parabola.
  • A vertex is also called an extreme point.
  • A parabola with a positive x2 coefficient (a>0) has 1 minimum point. The minimum is the point where the graph changes from decreasing to increasing. The shape of the parabola is convex, and it opens upward.
  • A parabola with a negative x2 coefficient (a<0) has 1 maximum point. The maximum is the point where the graph changes from increasing to decreasing. The shape of the parabola is concave, and it opens downward.
  • We can find the vertex of the parabola from its vertex function y=a(x-h)²+k, where the vertex coordinates are (h,k). Click here for more details about finding the vertex form of a quadratic function.

The graphs below show 2 functions: y=x²+2x+10 and y=-x²+2x+20.
The table near the graphs shows different (x,y) values that were taken to plot the graphs.
The vertex points are marked in red and green inside the table and on the graphs.

A vertex point of a parabola

The function y=x²+2x+10 has a vertex form y=(x+1)²+9. It has a positive x2 coefficient (a=1), a convex shape and a minimum point at (-1,9).

The function y=-x²+2x+20 has a vertex form y=-(x-1)²+21. It has a negative x2 coefficient (a=-1), a concave shape and a maximum point at (1,21).

A vertical symmetry of a parabola

All parabolas are symmetric: if we pass a vertical line through the vertex of the parabola, we will see that the right half is symmetric to the left half.

Note that since the parabola is symmetric, when we move the same distance from the x coordinate of the vertex, we get the same y values in both points.

For example: In the function y=x²+2x+10 the vertex is (-1,9). If we move to the right on the x axis to the point x=-1+1=0 we get y= 0²+2*0+10=10. If we move to the left on the x axis to the point x=-1-1=-2 we get y=-2²+2*-2+10=4-4+10=10. We got the same y value y=10.

The graphs below show 2 functions: y=x²+2x+10 and y=-x²+2x+20. One half of each parabola is colored in gray so you can see that the right half is symmetric to the left half.

The tables near the graphs show different (x,y) values that were taken to plot the graphs.

Some examples of vertex points are marked in colors inside the tables and on the graphs.

Vertical symmetry of parabolas images

The width of a parabola- wide parabolas

The x2 coefficient (a) determines the width of the parabola: The closer the value of x2 coefficient (a) to 0, the wider the parabola. 

If x2 coefficient (a) is very close to 0, the function of the parabola is almost like a straight line (x2*0=0), therefore the parabola is very wide.

The graphs below show 3 functions: y=2x+10, y=0.1x²+2x+10 and y=0.5x²+2x+10:

  • The function y=2x+10 is a linear function and its graph is a straight line.
  • The function y=0.1x²+2x+10 is a quadratic function in which we added to the linear function y=2x+10 a term of 0.1x². Since the coefficient of x² a=0.1 that is very close to 0, the parabola is very close to the graph of the straight line.
  • The function y=0.5x²+2x+10 is a quadratic function in which we added to the linear function y=2x+10 x² a term of 0.5x².

Since 0.5x²>0.1x² the distance from the straight line of the function y=0.5x²+2x+10 is bigger than distance of the function y=0.1x²+2x+10.

parabolas with small x square a coefficients

The table near the graphs shows different (x,y) values that were taken to plot the graphs.

The point (0,10) marked in green is the point of the intersection between the graphs, since in this point a=0 and the parabolas become straight lines.

A bigger x² coefficient (a) resulted in a bidder y value and a steeper parabola:
You can see from the table and the graph that for every point from x=0 the y values of the blue function with a=0.1 are smaller than the y values of the orange function with a=0.5.
The example is marked in red on the graphs and in the table. At the point x=-5 the blue function with a=0.1 equals to y=0.1*-5²+2*-5+10=0.1*25-10+10=2.5.
The orange function with a=0.5 equals to y=0.5*-5²+2*-5+10=0.5*25-10+10=12.5.
What made the difference between the y values is 0.5*25=12.5 compared to 0.1*25=2.5.
The black arrow on the graph shows the distance between 12.5-2.5.

The closer the value of x2 coefficient (a) to 0, the closer the parabola to the straight line:
You can see from the table and the graph that for every point from x=0 the y values of the blue function are closer to the y values of the grey straight line than the y values of the orange function with a=0.5. Therefore, the blue parabola with a=0.1 is closer to the grey straight line than the orange parabola with a=0.5.

The width of a parabola- narrow parabolas

A large absolute value of x2 coefficient (a) makes a steep and a narrow parabola. Parabolas with large negative or positive coefficients (absolute values) are far from 0 and therefore narrow. In other words: parabolas with large magnitudes (absolute values of the coefficients) of x2 (a) are steeper and narrower than parabolas with small magnitudes that are wide.

The graphs below show 4 functions with positive x2 coefficients: f(x)=x²+3x-10, f(x)=2x²+3x-10, y=3x²+3x-10 and y=4x²+3x-10. The only difference between the functions is the value of the x2 coefficient (a) that becomes bigger.

The table near the graph shows different (x,y) values that were taken to plot the graphs.

The width of a parabola- different x square positive coefficients graphs

The bigger the value of the x2 coefficient (a), the steeper is the parabola:

You can see from the table and the graph that for every point from x=0 the y values of the blue function with a=1 are smaller than the y values of the orange function with a=2 and so on for the functions with a=3 and a=4. Bigger y values result in bigger distances from a straight line (a=0) and a steeper parabola.

The example is marked in red on the graphs and in the table. At the point x=2 the blue function with a=1 equals to f(x)=x²+3x-10=2²+3*2-10=4+6-10=0.

The orange function with a=2 equals to f(x)=2x²+3x-10=2*2²+3*2-10=8+6-10=4.

The grey function with a=3 equals to f(x)=3x²+3x-10=3*2²+3*2-10=12+6-10=8.

The yellow function with a=4 equals to f(x)=4x²+3x-10=4*2²+3*2-10=16+6-10=12.

What made the difference between the y values is 1*2² compared to 2*2² and so on.

The black arrow on the graph shows the distance between the y values of 0, 4, 8 and 12.

The graphs below show 4 functions with negative x2 coefficients: f(x)=-x²+3x+10, f(x)=-2x²+3x+10, y=-3x²+3x+10 and y=-4x²+3x+10. The only difference between the functions is the value of the x2 coefficient (a) that becomes bigger in absolute value (magnitude): -1, -2, -3 and -4.

The table near the graph shows different (x,y) values that were taken to plot the graphs.

The width of a parabola- different x square negative coefficients graphs

The bigger the absolute value of the x2 (a) negative coefficient, the steeper the parabola:

You can see from the table and the graph that for every point from x=0 the y values of the orange function with a=1 are smaller (bigger in absolute value) than the y values of the blue function with a=2 and so on for the functions with a=3 and a=4. Smaller (bigger in absolute value) y values result in bigger distances from a straight line (a=0) and a steeper parabola.

Transforming graphs of quadratic functions:

Transforming parabolas can be done by changing their functions so that different types of changes cause different types of transformations. There are 3 types of parabola transformations:

  1. Parabola translation– moving the graph up, down, to the left or to the right without changing its width, so that the distance that each point moves is the same.
  2. Parabola stretching– changing the graph’s width, so that the distance that each point moves is not the same (leaving only the x intercepts coordinates the same).
  3. Parabola reflecting– reflecting the graph across the x axis or across the y axis.

Translating the parabola in the xy plane:

We can make 4 translations to the graph y=f(x).
Note that the distance that each point moves must be the same.

Shifting the parabola up or down:
In this case x values don’t change and y values change, therefore we need to change the value of y.

  1. Shifting the parabola up– the graph will move up so that the y value for every x will be bigger. This is possible by adding a constant c to the function, so that y=f(x)+c.
  2. Shifting the parabola down– the graph will move down so that the y value for every x will be smaller. This is possible by subtracting a constant c from the function, so that y=f(x)-c.

 Shifting the parabola to the left or to the right:
In this case y values don’t change and x values change, therefore we need to change the value of x.

  1. Shifting the parabola to the left– the graph will move to he left so that the x value for every y will be smaller. This is possible by adding a constant c to x, so that y=f(x+c).
  2. Shifting the parabola to the right– the graph will move to he right so that the x value for every y will be bigger. This is possible by subtracting a constant c from x, so that y=f(x-c).

Shifting the parabola up- adding a constant c to the function, so that y=f(x)+c:

We want the graph to move up, so that the y value for every x will be bigger (the distance that each point moves must be the same).

This is possible by adding a constant c to the function, so that y=f(x)+c. For example: shifting 2 points up will make the point (1,2) to be (1,4) and the point (3,4) to be (3,6).

Consider the following example:

Transform the graphs y=-x2+2x-5 and y=x²-4x+8 four points up.

The transformed graph of y=f(x) is y=f(x)+c therefore:
The graph y=-x2+2x-5 will become y=-x2+2x-5+4=-x2+2x-1
The graph y=x²-4x+8 will become y=x²-4x+8+4=x²-4x+12

The graphs below show the graphs of the given functions and their shifted graphs.

shifting a parabola up

The tables near the graphs show different (x,y) values that were taken to plot the graphs. The y values of each point in the shifted graphs are equal to the y values of the given graphs plus 4.

The red arrows near the graphs show the direction and the size of the movement- 4 points up.

As example, the points x=1 and x=3 are marked in brown and in purple in the tables and the graphs.

Shifting the parabola down- subtracting a constant c from the function, so that y=f(x)-c:

We want the graph to move down, so that the y value for every x will be smaller (the distance that each point moves must be the same).

This is possible by subtracting a constant c from the function, so that y=f(x)-c. For example: shifting 2 points down will make the point (1,2) to be (1,0) and the point (3,4) to be (3,2).

Consider the following example:

Transform the graphs y=-x2+2x-1 and y=x²-4x+12 four points down.

The transformed graph of y=f(x) is y=f(x)+c therefore:
The graph y=-x2+2x-1 will become y=-x2+2x-1-4=-x2+2x-5
The graph y=x²-4x+12 will become y=x²-4x+12-4=x²-4x+8

The graphs below show the graphs of the given functions and their shifted graphs.

shifting a parabola down

The tables near the graphs show different (x,y) values that were taken to plot the graphs. The y values of each point in the shifted graphs are equal to the y values of the given graphs minus 4.

The red arrows near the graphs show the direction and the size of the movement- 4 points down.

As example, the points x=2 and x=4 are marked in green and in purple in the tables and the graphs.

Shifting the parabola to the left - adding a constant c to the function, so that y=f(x+c):

The graph will move to he left so that the x value for every y will be smaller. Since y values do not change and x values change, this is possible by adding a constant c to x, so that y=f(x+c). For example: shifting 3 points to the left will change the point (1,2) to (-2,2) and the point (3,4) will change to (0,4).

Why do we add a value to x?
As we see in the example, for the function to move 3 units to the left each x value should become 3 units smaller. We want to plug x that is 3 units smaller into the transformed function and get the same y value as in the given function. In order to do that we need to add 3 units to x so it will become identical to the given x otherwise we cannot get the same y value (x-3+3=x). Therefore, if we take the transformed function y=f(x+3) and plug x values that are 3 units smaller we will get the original y values for every shifted x so that y=f(x+3-3)=f(x).

Consider the following example:

Transform the graphs y=-x2+2x-5 and y=x²-4x+8 three points to the left.

The transformed graph of y=f(x) is y=f(x)+c, therefore the graph y=-x2+2x-5 will become y=-(x+3)2+2(x+3)-5
Checking the functions:
Plugging x=1 in the original function: y=-x2+2x-5=-1+2-5=-4
Plugging x=1-3=-2 in the transformed function: y=-(x+3)2+2(x+3)-5=-(-2+3)2+2(-2+3)-5=-1+2-5=-4.
The y is the same (y=-4) in both functions and x moved 3 points to the left.

The transformed graph of y=f(x) is y=f(x)+c, therefore the graph y=x²-4x+8 will become y=(x+3)²-4(x+3)+8
Checking the functions:
Plugging x=1 in the original function: y=x²-4x+8=1-4+8=5
Plugging x=1-3=-2 in the transformed function: y=(x+3)²-4(x+3)+8=1-4+8=5
The y is the same (y=5) in both functions and x moved 3 points to the left.

The graphs below show the graphs of the given functions and their shifted graphs.

 

shifting a parabola to the left

The tables near the graphs show different (x,y) values that were taken to plot the graphs. The x values of each point in the shifted graphs are equal to the x values of the given graphs minus 3.

The red arrows near the graphs show the direction and the size of the movement- 3 points to the left.

As example, the points x=2 and x=-1 are marked in green and the point x=1 and x=4 are marked in purple in the tables and the graphs.

Shifting the parabola to the right - subtracting a constant c from the function, so that y=f(x-c):

The graph will move to he right so that the x value for every y will be bigger. Since y values don’t change and x values change, this is possible by subtracting a constant c from x, so that y=f(x-c). For example: shifting 3 points to the right will change the point (1,2) to (4,2) and the point (3,4) will change to (6,4).

Why do we subtract a value from x?
As we see in the example, for the function to move 3 units to the right each x value should become 3 units bigger. We want to plug x that is 3 units bigger into the transformed function and get the same y value as in the given function. In order to do that we need to subtract 3 units from x so it will become identical to the given x otherwise we can’t get the same y value (x+3-3=x). Therefore, if we take the transformed function y=f(x-3) and plug x values that are 3 units bigger we will get the original y values for every shifted x so that y=f(x-3+3)=f(x).

Consider the following example:

Transform the graphs y=-x2+2x-5 and y=x²-4x+8 three points to the right.

The transformed graph of y=f(x) is y=f(x)-c, therefore the graph y=-x2+2x-5 will become y=-(x-3)2+2(x-3)-5.
Checking the functions:
Plugging x=1 in the original function: y=-x2+2x-5=-1+2-5=-4
Plugging x=1+3=4 in the transformed function: y=-(x-3)2+2(x-3)-5=-(4-3)2+2(4-3)-5=-1+2-5=-4.
The y is the same (y=-4) in both functions and x moved 3 points to the right.

The transformed graph of y=f(x) is y=f(x)-c, therefore the graph y=x²-4x+8 will become y=(x-3)²-4(x-3)+8.
Checking the functions:
Plugging x=1 in the original function: y=x²-4x+8=1-4+8=5
Plugging x=1+3=4 in the transformed function: y=(x-3)²-4(x-3)+8=1-4+8=5
The y is the same (y=5) in both functions and x moved 3 points to the right.

The graphs below show the graphs of the given functions and their shifted graphs.

The tables near the graphs show different (x,y) values that were taken to plot the graphs. The x values of each point in the shifted graphs are equal to the x values of the given graphs plus 3.

The red arrows near the graphs show the direction and the size of the movement- 3 points to the right.

As example, the points x=1 and x=4 are marked in green and the point x=2 and x=5 are marked in purple in the tables and the graphs.

Stretching the parabola vertically in the xy plane:

Stretching by multiplying the function by a constant c>1

To stretch the graph vertically we need to multiply the function by a constant c>1. Stretching the graph y=f(x) vertically by a factor of c will give us a new graph y=f(x)*c.

If the graph is located above the x axis (all y values are positive), y values in the stretched graph will be bigger than in the original graph, therefore the stretched graph will move up above the original graph. Y values are bigger because a positive y multiplied by a number bigger than 1 will get us a bigger y in the stretched graph.

If the graph is located below the x axis (all y values are positive), y values in the stretched graph will be smaller than in the original graph, therefore the stretched graph will move down below the original graph. Y values are smaller because a negative y multiplied by a number bigger than 1 will get us a smaller y in the stretched graph.

Consider the following example:

Sretch the graphs y=-x2+2x-5 and y=x²-4x+8 vertically by a factor of 2.

The graphs below show the graphs of the given functions and their stretched graphs.

Stretching the parabola verticallywith no intercepts graph

The tables near the graphs show different (x,y) values that were taken to plot the graphs. The y values of each point in the stretched graphs are equal to the y values of the given graphs multiplied by 2.

The red arrows near the graphs show the direction of the movement. Note that the size of the movement changes between the points- bigger (in absolute values) y values result in bigger movements.

For example: the points x=0 and x=-2 are marked in green and purple in the tables and the graphs. The green point (0,-5) in the original graph stretched by 2 to the point (0,-10) in the stretched graph. The movement of the graph was -10 – -5= -10+5=-5 points down. The purple point (-2,-13) in the original graph stretched by 2 to the point (-2,-26) in the stretched graph. The movement of the graph was -26 – -13= -26+13=-13 points down.

If the graph is located below and above the x axis (the graph has 2 x intercepts), the movement of the stretched graph has 3 forms:

The intersection with the x axis where y=0 are the same points in the both graphs. This is because if we multiply y=0 in the original graph we get y=0 in the stretched graph.

The points above the x axis (where y is positive) will move up in the stretched graph so that the stretched graph will be above the original graph.

The points below the x axis (where y is negative) will move down in the stretched graph so that the stretched graph will be below the original graph.

Consider the following example:

Stretch the graphs y=-x2+2x-5 and y=x²-4x+8 by a factor of 2.

The graphs below show the graphs of the given functions and their stretched graphs.

The tables near the graphs show different (x,y) values that were taken to plot the graphs. The y values of each point in the stretched graphs are equal to the y values of the given graphs multiplied by 2.

The red arrows near the graphs show the direction of the movement. Note that the size of the movement changes between the points- bigger (in absolute values) y values result in bigger movements.

For example: the points x=-4 and x=0 are marked in green and purple in the tables and the graphs. The green point (0,14) in the original graph stretched by 2 to the point (0,28) in the stretched graph. The movement of the graph was 28-14=14 points up. The purple point (-4,-6) in the original graph stretched by 2 to the point (-4,-12) in the stretched graph. The movement of the graph was -12 – -6= -12+6=-6 points down.

Shrinking by multiplying the function by a constant 1>c>0

To shrink the graph vertically we need to multiply the function by a constant 1>c>0 Shrinking the graph y=f(x) vertically by a factor of c will give us a new graph y=f(x)*c. For example: the graph 0.5*f(x) is the graph f(x) shrunken by 2 (or stretched by 0.5).

Shrinking is the opposite to stretching:

For graphs above the x axis that have positive y values: if we multiply a positive y value by a number 1>c>0 we get a smaller y value, therefore the graph will move down.

For graphs below the x axis that have negative y values: if we multiply a negative y value by a number 1>c>0 we get a smaller negative y value, therefore the graph will move up.

If the graph is located below and above the x axis (the graph has 2 x intercepts), the movement of the stretched graph has 3 forms:
The intersection with the x axis where y=0 are the same points in the both graphs. This is because if we multiply y=0 in the original graph we get y=0 in the shrunken graph.
The points above the x axis (where y is positive) will move down in the shrunken graph so that the shrunken graph will be below the original graph.
The points below the x axis (where y is negative) will move up in the shrunken graph so that the shrunken graph will be above the original graph.

Consider the following example:

 

Shrink and stretch the graphs y=-x2+2x-5 and y=x²-4x+8 by a factor of 2.

The graphs below show the graphs of the given functions and their stretched and shrunken graphs.

Reflecting the parabola across the x axis and the y axis

A vertical reflection across the x axis

Reflecting the graph y=f(x) vertically will give us a new graph y=f(x)*-1.

Note that if we reflect the reflected graph again, we will get the original graph back, since y=f(x)*-1*-1=f(x). Meaning that the 2 graphs y=f(x) and y=-f(x) are reflection of each other.

Consider the following example:

Reflect across the x axis the parabola y=-x²+10x-21.

If the original graph is y=-x²+10x-21 then the reflected across the x axis graph will be
y=-1*(-x²+10x-21)
y=x²-10x+21

The graphs below show the graph of the given function and its reflected graph.

A vertical reflection of a parabola across the x axis

The black graph represents the original parabola and the orange graph represents the reflected parabola.  The green arrows near the graphs show the direction of the reflection.

The table near the graphs shows different (x,y) values that were taken to plot the graphs. The y values of each point in the reflected graph are equal to the y values of the given graph multiplied by -1. As example, the point x=5 is marked in red in the tables and the graphs.

Note that the black graph is also a vertical reflection of the orange graph:
The orange graph: y=x²-10x+21
The reflection of the orange graph is y=-1(x²-10x+21)=-x²+10x-21, this is the black graph.

A horizontal reflection across the y axis

A horizontal reflection reflects a graph horizontally across the y axis, so that the reflected graph is a mirror image of the original graph about the y axis (the mirroring is on the left and on the right of the y axis).

In order to move the graph across the y axis we need to change the sign of x coordinate for every y coordinate. Meaning that in order to make a reflection across the y axis the y coordinate does not change, and the x coordinate changes its sign.  For example: If the original point is (1,2) the reflected across the x axis point will be (-1, 2).

Reflecting the graph y=f(x) horizontally will give us a new graph y=f(x*-1).

Note that if we reflect the reflected graph again, we will get the original graph back, since y=f(x*-1*-1)=f(x). Meaning that the 2 graphs y=f(x) and y=f(-x) are reflection of each other.

Consider the following example:

Reflect across the x axis the parabola y=-x²+10x-21.

If the original graph is y=-x²+10x-21 then the reflected across the y axis graph will be
y=-(x*-1)²+10x*-1-21
y=-x²-10x-21

The graphs below show the graph of the given function and its reflected graph.

A horizontal reflection graphs

The black graph represents the original parabola and the green graph represents the reflected parabola.  The blue arrows near the graphs show the direction of the reflection.

The table near the graphs shows different (x,y) values that were taken to plot the graphs. The x values of each point in the reflected graph are equal to the y values of the given graph multiplied by -1. As example, the point y=4 is marked in red in the tables and the graphs.

Note that the black graph is also a horizontal reflection of the green graph:

The green graph is y=-x²-10x-21.

The reflection of the green graph: y=-(x*-1)²-10x*-1-21=-x²+10x-21 this is the black graph. 

A horizontal reflection across the y axis versus a vertical reflection across the x axis

The graphs below combine horizontal and vertical reflections of the graph of y=-x²+10x-21 explained earlier.

A horizontal reflection versus a vertical reflection

The black graph represents the original parabola.
The green graph represents the horizontal reflection of the parabola across the y axis.
The orange graph represents the vertical reflection of the parabola across the x axis.

The table near the graphs shows different (x,y) values that were taken to plot the graphs. As example, the point (5,4) and its reflected points are marked in red in the tables and the graphs. The point (5,4) was first reflected across the x axis to the point (5,-4) in the orange graph and then it was reflected across the y axis to the point (-5,4) in the green graph.

SAT worksheet- Graphing quadratic functions

Function notation

sat function notation

Function notation on the SAT

SAT Subscore: Passport to Advanced Mathematics

.A function is a formula that represents relationship between input variable x (also called independent variable) and output variable f(x) (also called dependent variable). For example: f(x)=2x+5.

In function notation we evaluate the function, meaning finding the output of the function f(x) for a given input x. The input can be numeric (a number), an expression or another function.

An example for a function notation with a numeric input: the value of a function f(x)= 2x+5 when x=2 is f(2)=2x+5=2*2+5=9.
An example for a function notation with an  expression input: the value of a function f(x)= 2x+5 when f(x-1) is f(x-1)=2(x-1)+5=2x-2+5=2x+3.
An example for a function notation with a function g(x) as an input in another function f(x): the value of a function f(x)= 2g(x)+5 when g(x)=2x is f(g(x))=2(2x)+5=4x+5.

Representing values in a table instead of evaluating a function:
Instead of evaluating a function using its formula we can receive a table providing pairs of inputs and outputs with or without seeing the formula that stands behind the table.

For example: Consider the table below.

x              f(x)
2               9
t+1           2t+7

The formula that is behind the table is f(x)= 2x+5.

Evaluating a function at a numeric input

Since the input is a number, we can calculate the output of a function as a number.

There are 2 steps for evaluating a function using its formula:
Step 1: Plugging the input value instead of the input variable.
Step 2: Calculating the output of the function.

Consider the following example:

What is the value of the function f(x)=3-x when x=3?

Step 1: Plugging the input value 3 instead of the input variable x:
Remember the negative exponents formula x-n=1/xn
f(x)=3-x

Step 2: Calculating the output of the function:
f(3)=3-3=1/33=1/(3*3*3)=1/27

Consider the following example:

The function f(x) is f(x)=(x-1)(x+2).

What is the value of f(2)-f(1)?

f(2)=(x-1)(x+2)=1*4=4

f(1)=0

f(2)-f(1)=4-0=4

Evaluating a function with an expression input

In these questions the input is an expression instead of a single number, therefore in step 1 we need the plug the expression into the function the same way we plug the number.

Consider the following example:

If f(x)=(x+2)2 what is f(x-1)?

Step 1- Plugging the input expression x-1 instead of the input variable x:
f(x-1)=(x-1+2)2=(x+1)2

Step 2: Calculating the output of the function:
Remember square of sum formula: (a+b)2=a2+b2+2ab
(x+1)2=x2+1+2x=x2+2x+1

Evaluating a function output using a table

In these questions we receive a table providing pairs of inputs and outputs without seeing the formula that stands behind the table.

There are 2 steps for evaluating a function output using a table:

Step 1: Finding the given input value in the input column of the table (the heading of the input column is x).

Step 2: Finding the output value that corresponds to the input in the output column of the table (the heading of the output column is f(x)).

Consider the following example:

The table below contains 3 input-output pairs for a function.

x    f(x)
-2    2
0    -2
1    -1

What is the value of f(-2)?

Step 1: Finding the input -2 in the input column of the table (the left column x)

The input -2 is written in the first row.

Step 2: Finding the output value that corresponds to the input -3 it the output column of the table (the right column f(x)).

Since we found that the given input is in the first row of the table, we need to look for the corresponding output in the first row of the table. The output that is written in the first row and a second column f(x) of the table is 2. Therefore f(-2)=2.

Note that you need to look for the input -2 in the left x column and not in the right f(x) column. If you confuse between the columns you will get a wrong answer that is 0.  

Composing 2 functions into a composite function

A composite function is a function that depends on another function as an input instead of a single variable x. For example: f(g(x)) is a composite function in which g(x) is substituted for x in f(x).

Note that f(g(x)) in different from g(f(x)). In order to evaluate f(g(x)) we need to substitute g(x) in f(x) while in order to evaluate g(f(x)) we need to substitute f(x) in g(x).

The steps for writing a composite function f(g(x)):

Step 1- evaluate the value of the outer function f(x) with the input x=g(x) by plugging the function of g(x) in f(x) instead of x.

Step 2- simplify the outer function: open brackets and combine like terms.  

 

Consider the following example:

if f(x)=2x+3 and g(x)=4x+1

What is the value of f(g(x))?
What is the value of g(f(x))?

f(g(x))=2(4x+1)+3=8x+2+3=8x+5 g(f(x))=4(2x+3)+1=8x+12+1=8x+13

Evaluating composite functions at a given input value

In these questions we are given the value of x, therefore we can calculate the numeric value of the composite function.

The steps for evaluating a composite function f(g(x)) at a given input value:

Step 1- calculate the value of the inner function g(x) with the given input x.

Step 2- calculate the value of the outer function f(x) with the input x=g(x) that was calculated in step 1.

Another way is to compose f(g(x)) first and then plug the given x into the composed function. This option can take longer therefore is less recommended.

 

Consider the following example:

if g(x)=2x+3 and f(x)=4x+1 what is the value of f(g(-1))?

We need to calculate the inner function g(x) first and then plug the answer into the outer function f(x).

Step 1– calculate the value of the inner function g(x):
g(x)=2x+3
g(-1)=2*-1+3=1

Step 2– calculate the value of the outer function f(x):
f(x)=4x+1
f(1)=4*1+1=5

Another way is to compose f(g(x)) first and then plug the given x into the composed function:
f(g(x))=4(2x+3)+1=8x+13
f(g(-1))=8*-1+13=-8+13=5

The answer is the same in both ways.

Evaluating composite functions using tables

The table of a composite function has 3 columns instead of 2:
input column– x
output column of the function f(x) when we plug x into f(x)
output column of the function g(x) when we plug x into g(x)

We need to find the relevant values in the table: the given input is used as an input of the inner function while the output of the inner function from the table is used as an input for the outer function. Meaning that we must use the table twice with 2 different inputs and 2 different outputs.

The steps for evaluating composite function f(g(x)) using tables:
Step 1: Finding the given input value in the input column x.
Step 2: Finding the output value that corresponds to the given input x in the column of the inner function (the heading is g(x)).
Step 3: Finding the output value of the inner function g(x) that was calculated in step 2 in the input column x.
Step 4: Finding the output value that corresponds to the output value of the inner function g(x) in the column of the outer function (the heading is f(x)).

Consider the following example:

The table below provides the values of functions f(x) and g(x) at several values of x.

Calculate is the value of f(g(1)), f(g(0)), g(f(-1)) and g(f(-2)).

 x    g(x)   f(x)
-2      4       0
-1      1       2
 0      0       4
 1      1       6
 2      4       8

Calculating the value of f(g(1)):

The steps are marked in the headings and in the table below:

Step 1: Finding the given input 1 in the input column x
The row is the fourth row.

Step 2: Finding the output value that corresponds to the input 1 in the column of the inner function g(x).
The column is the second column and the output is g(1) is 1.

Step 3: Finding the output 1  in the input column x
The row is the fourth row.

Step 4: Finding the output value in the column of the outer function f(x)
The column of the outer function f(x) is the third column. The input of the inner function g(1)=1 is located in the fourth row therefore the output of the outer function f(x) is 6.

The answer is f(g(1))=6.

function notation example 1 with table

Calculating the value of f(g(0)):

The steps are marked in the headings and in the table below:

Step 1: Finding the given input 0 in the input column x
The row is the third row.

Step 2: Finding the output value that corresponds to the input 0 in the column of the inner function g(x).
The column is the second column and the output is g(0) is 0.

Step 3: Finding the output 0  in the input column x
The row is the third row.

Step 4: Finding the output value in the column of the outer function f(x)
The column of the outer function f(x) is the third column. The input of the inner function g(0)=0 is located in the third row therefore the output of the outer function f(x) is 4.

The answer is f(g(0))=4.

function notation example 4 with table

Calculating g(f(-1)):

The steps are marked in the headings and in the table below:

Step 1: Finding the given input -1 in the input column x

The row is the second row.

Step 2: Finding the output value that corresponds to the input -1 in the column of the inner function f(x).

The column is the third column and the output is f(-1) is 2.

Step 3: Finding the output 2 in the input column x

The row is the fifth row.

Step 4: Finding the output value in the column of the outer function g(x)

The column of the outer function g(x) is the second column. The input of the inner function f(-1)=2 is located in the fifth row therefore the output of the outer function f(x) is 4.

The answer is g(f(-1))=4.

function notation example 3 with table

Calculating g(f(-2)):

The steps are marked in the headings and in the table below:

Step 1: Finding the given input -2 in the input column x

The row is the first row.

Step 2: Finding the output value that corresponds to the input -1 in the column of the inner function f(x).

The column is the third column and the output is f(-2) is 0.

Step 3: Finding the output 0 in the input column x

The row is the third row.

Step 4: Finding the output value in the column of the outer function g(x)

The column of the outer function g(x) is the second column. The input of the inner function f(-2)=0 is located in the third row therefore the output of the outer function f(x) is 0.

The answer is g(f(-2))=0.

function notation example 4 with table

Isolating variables

Isolating variables questions on the SAT

SAT Subscore: Passport to Advanced Mathematics

Isolated variable is a variable that is written alone on one side of the equation and the second side of the equation contains other variables. For example: In the equation y=mx+b the variable y is isolated.

Insulating versus solving: We can isolate any variable of the equation by rearranging the equation according to the same rules we use to solve it. These are the rules for maintaining equality: we can add or subtract the same value from each side of the equation and we can divide or multiply each side of the equation by a same value.

Note that the isolated variable will be equal to an expression containing other variables and not a numeric solution like after solving an equation.

Isolating variables questions may ask you to isolate variables in different types of equations, such as linear equations or quadratic equations.

If you are not familiar with solving an equation topic visit the page about linear equations before learning this topic.

Isolating variables in linear equations

You need to follow the rules for maintaining equality, in addition you may need to use distributing and combining like terms. If you need to learn more, see linear equations topic.

Consider the following example:

Slope intercept form of a linear function is given by the equation y=3x+7. What is x in terms of y?

y=3x+7
y-7=3x+7-7
y-7=3x
x=(y-7)/3

Now we can calculate x coordinate given y coordinate area. For example: if y coordinate is 16 then x coordinate is x=(16-7)/3=3.

Checking the answer: 
y=3x+7
y=3((y-7)/3)+7
y=y-7+7
y=y

Consider the following example:

5a2-7b=3a2-3b+5

What is b in terms of a?

5a2-7b=3a2-3b+5
-3b+7b=5a2-3a2-5
4b=2a2-5
b=(2a2-5)/4

Checking the answer:
5a2-7b=3a2-3b+5
5a2-7(2a2-5)/4=3a2-3(2a2-5)/4+5

Multiply each part of the equation by 4:
20a2-7(2a2-5) =12a2-3(2a2-5)+20
20a2-14a2+35=12a2-6a2+15+20
6a2+35=6a2+35

Consider the following example:

The plant uses function c defined by c(n)=100,000/n+5 to calculate the cost c(n), in dollars, of producing one product. The variable production cost is 5 dollars, the fixed cost of the plant is 100,000 dollars and n is defined as the quantity of the products.

Rewrite n in terms of c.

c=100,000/n+5
c-5=100,000/n+5-5
c-5=100,000/n
n(c-5)=100,000n/n
n(c-5)=100,000
n(c-5)/(c-5)=100,000/(c-5)
n=100,000/(c-5)

Now we can calculate the number of units that were produced given the production cost of a product. For example: if the production cost of a product is 10 dollars, what is the number of products that the plant produced?
n=100,000/(c-5)
n=100,000/(10-5)
n=100,000/5=20,000

Checking the answer:
c=100,000/n+5
c=100,000/(100,000/(c-5))+5
c=100,000*((c-5)/100,000)+5
c=c-5+5
c=c

Isolating variables in quadratic equations without x term

Since we have only x2 term and there is no x term in the equation, we can isolate x

Step 1: Isolate x2 with the rules for maintaining equality, in addition you may need to use distributing and combining like terms. If you need to learn more, see linear equations topic.

Step 2: Take a square root from each side of the equation. If you need to learn more, see quadratic equations topic.

Consider the following example:

The area of a square is the product of the length l of each side with itself A=l2.

What is l in terms of A?

Step 2: Take a square root of both sides of the equation:
A=l2
√ A=√ l2
√ A=l
l=√ A

Now we can calculate the length of each side of a square given its area. For example: if the area is 49 then the length of its side is √ 49=7.

Checking the answer:
A=l2
A=√ A 2
A=A

Consider the following example:

5a2-7b=3a2-3b+5

What is a equal to in terms of b?

Step 1: Isolate x
5a2-7b=a2-3b+5
5a2-a2=-3b+7b+5
4a2=4b+5
4a2/2=(4b+5)/2
a2=(4b+5)/4

Step 2: Take a square root from each side of the equation
a2=(4b+5)/4
√a2=√(4b+5)/√4
a=√(4b+5)/2

Checking the answer:
5a2-7b=a2-3b+5
5(√(4b+5)/2)2-7b=(√(4b+5)/2)2-3b+5
(5(4b+5))/4 -7b=(4b+5)/4-3b+5

Multiply each part of the equation by 4:
5(4b+5)-28b=4b+5-12b+20
20b+25-28b=4b+5-12b+20
-8b+25=-8b+25

Quadratic equations and quadratic functions on the SAT

Quadratic equations and quadratic functions on the SAT

SAT Subscore: Passport to Advanced Mathematics

quadratic equations and functions formulas sheets

A quadratic expression (in Latin quadratus means squared) is an expression that has a squared term (a variable multiplied by itself) and is usually written in a standard form ax2+bx+c where a≠0 (x is a variable and a, b and c are constants).

Quadratic functions and equations questions may ask you to rewrite a quadratic function to one of its 3 forms or to solve a quadratic equation using a formula. Rewriting the quadratic function allows to display its different features: the coordinates of the vertex of the parabola and the x intercepts.

Finding the function of the parabola from its graph topic includes questions about finding the standard form, the quadratic form and the vertex form equations using the vertex coordinates and the x and y intercepts coordinates from the graph of the function.

Other questions include two types of word problems with quadratic functions:  calculating the area of a rectangle problems and calculating height versus time problems. The skills on this subject require writing a quadratic function from a word problem and solving a given quadratic function according to a word problem.

There are 3 forms for writing a quadratic function:

Standard form of a quadratic function:
y=ax2+bx+c

For example: y=x2-2x-8

Factored form of a quadratic function:
y=(ax+c)(bx+d)

For example: y=(x+2)(x-4)

Vertex form of a quadratic function:
y=a(x-h)2+k

For example: y=(x-1)2-9

The standard form is showing the y intercept of the graph, it is written as y=ax2+bx+c.
For example: y=x2-2x-8 is a standard form a quadratic function.

The factored form is showing the x intercepts of the graph, it is written as y=(ax+c)(bx+d). For example: The factored form of the function given above is y=(x+2)(x-4), therefore we know that the function intercepts the x axis at the points x=-2 and x=4 (y=0).
Note that you can open brackets of y=(x+2)(x-4) and get its standard form y=x2-2x-8.

The vertex form is showing the coordinates of the vertex of the parabola (the main point of a parabola is its vertex), it is written as y=a(x-h)2+k.
For example: The vertex form of the function given above is y=(x-1)2-9, therefore we know that the function has a vertex at the point x=1; y=-9.

Note that:
1. a≠0, otherwise the expression will be linear and not quadratic in a form of y=bx+c.
2. X is a variable and a, b, c, d, h and k are constants.
3. The letters a, b and c are used to represent constant values separately in every formula. Meaning that the constants a, b and c in the formula y=ax2+bx+c are different from the constants a, b and c in the formula y=(ax+c)(bx+d) and are different from the constant a in the formula y=a(x-h)2+k.

Finding the factored form of a quadratic function questions

The types of questions on this subject can be:

1. We are given a quadratic function in a standard form as y=ax2+bx+c and we need to write it in a factored form as y=(ax+c)(bx+d).

2. We are given a quadratic function in a standard form as y=ax2+bx+c and we are asked to find the values of the x intercepts of the graph.

How can we find the factored form of a quadratic function with a=1?

In these questions we are given a simplified quadratic function y=1*x2+bx+c equal to y=x2+bx+c (a=1) and we are asked to find the factored form of this simplified function.

If we are  given the standard form y=x2+bx+c (the coefficient of x2 is 1), the factored form will be y=(x+c)(x+d) (the coefficients of x are also 1). This is why:

When we open brackets in the factored form, we multiply x by x to get x2 (this is the first expression it the FOIL formula- ax*bx). If the coefficients of the x variables in the factored form are different from 1 we can’t get 1 as a coefficient of x2 in the standard form. Therefore if we are given a structure of the standard form as y=x2+bx+c (meaning that a=1) then the structure of the factored form must be y=(x+c)(x+d) (meaning that a=1 and b=1).

For example: (x+2)(x-1)=x2-x+2x-2= x2+x-2. The coefficients of x in both parts of the factored form are 1 and the coefficient of x2 in the standard form is 1.

For example: (2x+2)(3x-1)=6x2-2x+6x-2= x2+4x-2. The coefficients of x in the factored form are a=2; b=3 and the coefficient of x2 in the standard form is a=6 (and not 1).

How to find the factored form given the standard form?

The factored form with FOIL formula is y=(x+c)(x+d)= x2+dx+cx+cd= x2+(c+d)x+cd.
The standard form is x2+bx+c.

We see that the constant c in the standard form is calculated from the multiplication of the constants c and d of the factored form. We also see that the constant b in the standard form is calculated from the sum of the constants c and d in the factored form. Therefore, if we are given the standard form, we know its b and c values and we can find the values of c and d of the factored form.

Consider this example:

Find the factored form of the function y=x2-2x-8.

In this example we have a standard form of y=x2-2x-8 (a=1, b=-2, c=-8).

Since a=1 in the standard form we know that a=1 and b=1 in the factored form. Therefore the factored form is y=(x+c)(x+d).

We know that c=-8 is equal to c*d in the factored form.

We know that b=-2 in the standard form is equal to c+d in the factored form.

What are the 2 numbers whose product is equal to -8 and whose sum is equal to -2? The numbers are -4 and 2. Therefore the factored form is (x+2)(x-4).

Checking by opening brackets with FOIL formula:
(x+2)(x-4)= x2-4x+2x-8= x2-2x-8. We see that the factored function is the same as the given standard function.

Finding the factored form of a quadratic function with a≠1: factoring out a common factor

We can use this method only if we can factor out a common factor so that we will be left with a function in which the coefficient of  x2 is 1.

Consider the following example:

Find the factored form of the function y=2x2-6x-8.

Note that the coefficient of x2 is not 1, it is 2. In this example we can factor out a common factor and stay with 1 as x2 coefficient:
y=2x2-6x-8=2(x2-3x-4)

Now we have a new standard form of y=x2-3x-4 (a=1, b=-3, c=-4).

Since a=1 in the new standard form we know that a=1 and b=1 in the factored form. Therefore the factored form is y=(x+c)(x+d).

We know that c=-4 is equal to c*d in the factored form.

We know that b=-3 in the standard form is equal to c+d in the factored form.

What are the 2 numbers whose product is equal to -4 and whose sum is equal to -3? The numbers are -4 and 1. Therefore the factored form is (x-4)(x+1).

Remember that the functions was y=2x2-6x-8=2(x2-3x-4) and we found the factored function for x2-3x-4. Therefore the factored function for the original function is 2(x-4)(x+1).

Check by opening brackets with FOIL formula:
2(x-4)(x+1)= 2(x2+x-4x-4)=2(x2-3x-4)=2x2-6x-8. We see that the factored function is the same as the given standard function.

Finding the factored form of a quadratic function with a≠1 without factoring out a common factor

These are the most difficult questions, since we can’t factor out a common factor and be left with a function in which the coefficient of  x2 is 1. Since the x2 coefficient is no longer 1 it is not enough to find 2 numbers whose sum equals b and whose multiplication equals c. We need to check the options for 2 numbers that also considering the x2 coefficient that is a. 

If you are given the answers in multiple choice question you can check every answer by opening brackets with FOIL formula instead of solving the given function.

Consider the following example:

Find the factored form of the function y=4x2+25x+6.

Since the x2 coefficient is no longer 1 it is not enough to find 2 numbers whose sum is 6 and whose multiplication is 25. We need to check the options for 2 numbers that also considering the x2 coefficient that is 4.

We know that there are 2 options for the factored form so that x2 coefficient will be 4:
Option 1 is (4x±a )(x±b)   In this option multiplying the first numbers will get 4*1*x2=4x2.
In this option the coefficient of x is calculated by 4bx+ax, meaning x(4b+a) which is 4 times number b and number a.

Option 2 is (2x±a )(2x±b)  In this option multiplying the first numbers will get 2*2*x2=4x2.
In this option the coefficient of x is calculated by 2bx+2ax, meaning x(ab+2a) which is 2 times number b and 2 times number a.

The question asked should be: what are the numbers whose multiplication is 6 and one the following:
4 times number b and number a is 25: 4b+a=25
2 times number b and 2 times number a is 25: 2a+2b=25

The numbers whose multiplication is 6 are 1 and 6 or 2 and 3. For each one of these two options we also need to look at the second criteria and find a match.

If the numbers are 6 and 1 (6*1=6):
4 times number b and number a is 25: 4b+a=25

This is possible since 6*4+1=25, therefore the numbers are 6 and 1 and the coefficients of x2 are 6 and 4. The factored form is (4x+1)(x+6).

Checking the answer by opening brackets with FOIL formula: (4x+1)(x+6)=4x2+24x+x+6=4x2+25x+6.

Let’s check the other option: 2 times number b and 2 times number a is 25: 2a+2b=25
We can’t get this with the numbers 1 and 6 since 2*1+2*6=14 and not 25.

If the numbers are 2 and 3 (2*3=6):
4 times number b and number a is 25: 4b+a=25
We can’t get this with the numbers 2 and 3 since 2*1+3*4=16 and not 25, also 3*1+2*4=11 and not 25.

Let’s check the other option: 2 times number b and 2 times number a is 25: 2a+2b=25
We can’t get this with the numbers 2 and 3 since 2*2+3*2=10 and not 25.

How can we find the factored form of a quadratic function using special factoring?

In order to use special factoring, we need to remember the following formulas:

Square of sum formula:
(a+b)2=a2+b2+2ab

For example: (2x+5)2=4x2+25+20x

Square of difference formula:
(a-b)2=a2+b2-2ab

For example: (2x-5)2=4x2+25-20x

Difference of squares formula:
a2-b2=(a+b)(a-b)

For example: 4x2-25=(2x+5)(2x-5)

In this solving method we will be given the result of one of these 3 formulas so we can factor it immediately using one of the formulas. In order to do it we need to look for perfect squares.
Perfect squares are integers which square is an integer. Note that the numbers a2 and b2 are perfect squares since their squares a and b are also integers.

Steps for finding the factored form of a quadratic function using special factoring:
1. Factor out common factors if it will help you to get perfect squares for the next step.
2. Find the components of the functions a2+b2±2ab or a2-b2 and do special factoring using the formulas:

When we see a sum of 2 numbers that are perfect squares, we know that they can be a2 and b2 and we need to see if we also have the third part of the formula which is 2ab or -2ab to complete the formula a2+b2+2ab=(a+b)2 or a2+b2-2ab =(a-b)2.

When we see a difference of 2 numbers that are perfect squares, we know that they can be a2 and b2 and we factor the formula a2-b2=(a+b)(a-b).
 

Consider the following example:

What is the factored form of a function f(x)=8x4+2x2+8x3?

Step 1: Factor out common factors:
We can factor out 2 as a common factor: 8x4+2x2+8x3=2(4x4+x2+4x3).

Step 2: We see that 4x4 is a perfect square since its square root is an integer: √4x4=2x2.
We see that x2 is a perfect square since its square root is an integer: √x2=x.
We see that the multiplication of the 2 perfect squares by 2 gives us the third number 2x2*x*2=4x3.

One of the numbers is negative so we know that we can do a special factoring using the formula a2-b2=(a+b)(a-b) getting  2(4x4+x2+4x3)=2(2x2+x)2.

Note that we can’t solve without factoring out 2 as a common factor since 8x4 and 2x2 are not perfect squares. In this case we must factor out a common factor before the perfect factoring.

Consider the following example:

What is the factored form of a function f(x)=16x4-4x2?

Step 1: Factor out common factors:
We can factor out 4 as a common factor: 16x4-4x2=4(4x4-x2).

Step 2: We see that 4x4 is a perfect square since its square root is an integer: √4x4=2x2.
We see that x2 is a perfect square since its square root is an integer: √x2=x.

One of the numbers is negative and so we know that we can do a special factoring using the formula a2-b2=(a+b)(a-b) getting  4(4x2-x2)=4(2x2+x)(2x2-x).

Note that factoring out 2 as a common factor will get numbers 8x4 and 2x2 that are not perfect squares 16x4-4x2=2(8x4-2x2), therefore it is not suitable because it makes the next step impossible.

Note that we can solve without factoring out 4 as a common factor since 16x4 and 4x2 are also perfect squares. In this case we will need to factor out a common factor after the perfect factoring.
16x4-4x2=(4x2+2x)(4x2-2x)=2(2x2+x)*2(2x2-x)=2*2*(2x2+x)(2x2-x)=4(2x2+x)(2x2-x).

Consider the following example:

What is the factored form of a function f(x)=20x2-5?

Step 1: Factor out common factors:
We can factor out 5 as a common factor: 20x2-5=5(4x2-1).

Step 2: We see that 4x2 is a perfect square since its square root is an integer: √4x2=2x.
We see that 1 is a perfect square since its square root is an integer: √1=1.

One of the numbers is negative and so we know that we can do a special factoring using the formula a2-b2=(a+b)(a-b) getting  5(4x2-1)=5(2x+1)(2x-1).

Note that we can’t solve without factoring out 5 as a common factor since 20x2 and 5 are not perfect squares. In this case we must factor out a common factor before the perfect factoring.

How can we find the x intercepts of a quadratic function using its factored form?

We know that the y values of the interception points with the x axis are y=0. Therefore we know that at the factored form equation at the x axis interception points is (x+c)(x+d)=0.

The zero product property states that if the multiplication of 2 numbers is equal to 0 then one of the numbers must be equal to 0. In the factored formula equation that means that x+c= 0 or x+d=0. Since c and d are constants that we know we can see the x values from the factored formula. For example: in the function y=(x+2)(x-4)=0, the intercepts with the x axis are at the points x=-2 and x=4.

Consider the following example:

Find the x intercepts of the function f(x)=x2+3x-10.

Step 1: we need to find the factored form of the function. Since the standard form the is simplified to y=x2+bx+c (a=1) we know that the factored form will be (x+c)(x+d) (so that a=1, b=1).
The coefficient -10 in the standard formula is calculated from the multiplication of c and d in the factored formula.
The coefficient 3 in the standard formula is calculated from the sum of c and d in the factored formula.

What are the 2 numbers whose product is equal to -10 and whose sum is equal to 3? The numbers are -2 and 5. Therefore the factored form is (x-2)(x+5).

Check by opening brackets with FOIL formula:
(x-2)(x+5)= x2+5x-2x-10= x2-3x-10. We see that the factored function is the same as the given standard function.

Step2: We need to solve the equation (x-2)(x+5)=0.
According to the zero product property we know that the function has 2 x intercepts x=2 and x=-5.

The following graph presents the graph of the function with its intercepts. 

graph of the factored form x intercepts

Finding the vertex form of a quadratic function questions

The types of questions on this subject can be:
1. We are given a quadratic function in a standard form as y=ax2+bx+c and we need to write it in a vertex form as y=a(x-h)2+k.
2. We are given a quadratic function in a standard form as y=ax2+bx+c and we are asked to find the values of the vertex of the graph.

How can we find the vertex form of a quadratic function?
If we look at the vertex formula y=a(x-h)2+k we see the quadratic expression (x-h)2.
(x-h)2=(x-h)(x-h)

Opening brackets with FOIL formula will give us:
(x-h)(x-h)=x2-hx-hx+h2= x2-2hx+h2 (h is constant).

Note that the form of x2-2hx+h2 is structured like the standard form y=ax2+bx+c. There is a squared variable, a constant and a multiplication of a constant and a variable.

How can we find a vertex form from a standard form? We will rewrite the standard form to fit to the vertex form.

Finding the vertex form of a function with a minus sign in the x coefficient

Consider the following example:

The standard form is y= x22x-8. What is the vertex form of the function?

Look at the expression x2-2x-8. What do we need to do in order to write it in a form of a(x-h)2= x2-2hx+h2?

Making x2-2x-8 to be x2-2hx+h2:
We have x2 as x2
We have -2x as -2hx
We have -8 as h2

Therefore, we see that if h=1 well have an equality in x2 and -2x and what is left is only h2. his supposed to be 1 and it is -8.

That means we have x2-2x+1 and we need minus 9 to finish the standard form:
x2-2x-8=x2-2x+1-9

Now we can write x2-2x+1 as a square equation:
x2-2x+1=(x-1)2

Writing the standard form again will give us the vertex form:
x2-2x-8=x2-2x+1-9=(x-1)2-9

The vertex formula is y=a(x-h)2+k therefore the constants are a=1, h=1 and k=-9.

Checking the answer: we can open brackets of the vertex form to see that we get the standard form:
(x-1)2-9=(x-1)(x-1)-9=x2-x-x+1-9= x2-2x-8.

Finding the vertex form of a function with a plus sign in the x coefficient

Consider the following example:

The standard form is y=x2+2x+10. What is the vertex form of the function?

Look at the expression x2+2x+10, since we have a plus sign before 2x we will write the vertex form as a(x+h)2=x2+2hx+h2 and deal with the sign later.

Making x2+2x+10 to be x2+2hx+h2:
We have x2 as x2
We have 2x as 2hx
We have 10 as h2

Therefore, we see that if h=1 well have an equality in x2 and 2x and what is left is only h2. his supposed to be 1 and it is 10.

That means we have x2+2x+1 and we need 9 to finish the standard form:
x2+2x+10=x2+2x+1+9

Now we can write x2+2x+1 as a square equation:
x2+2x+1=(x+1)2

Writing the standard form again will give us the vertex form:
x2+2x+10=x2+2x+1+9=(x+1)2+9
The vertex formula is y=a(x-h)2+k therefore the constants are a=1, h=-1 and k=9.

Note that in order to get x+1 from the formula x-h we need h to be h=-1 and not h=1, meaning that x-h=x – -1=x+1.

Checking the answer: we can open brackets of the vertex form to see that we get the standard form:
(x+1)2+9=(x+1)(x+1)+9=x2+x+x+1+9= x2+2x+10.

How can we find the vertex values of a quadratic function using its vertex form?

The vertex form y=a(x-h)2+k has 2 expressions a(x-h)2 and k. Note that the expression in the brackets has a quadratic exponent therefore it is positive for every x.

The x coordinate of the vertex is equal to h and the y coordinate of the vertex is equal to k. This is why:

If a is positive then the function has a minimum point. At the minimum point, the y value of the function must be the smallest. If the first expression a(x-h)is not equal to 0, it will cause the y value to be bigger and we want it to be the smallest, therefore at the minimum point the first expression a(x-h)2 must be equal to 0 so that what is left is only the constant k. What are the conditions for a(x-h)2=0? We need x to be equal to h and then y will be equal to k.

If a is negative then the function has a maximum point. At the maximum point, the y value of the function must be the biggest. Note that since a is negative the first expression a(x-h) 2  will be always negative. If the first expression a(x-h)2 is not equal to 0, it will cause the y value to be smaller and we want it to be the biggest, therefore at the maximum point the first expression a(x-h)2 must be equal to 0 so that what is left is only the constant k. What are the conditions for a(x-h)2=0? We need x to be equal to h and then y will be equal to k.

Consider the following example:

What are the coordinates of the vertex in the function (x-1)2-9?

Since the coefficient of x2 is positive, we know that the function has a minimum point.
In order for y to be the smallest possible we need the expression (x-1)2  to be 0.
This is possible when x=1. Therefore, at the vertex point x=h=1 and y=k=-9.

The following graph presents the graph of the function with its vertex point. 

vertex form graph

Consider the following example:

What are the coordinates of the vertex in the function (x+1)2+9?

Since the coefficient of x2 is positive, we know that the function has a minimum point. In order for y to be the smallest possible we need the expression (x+1)2  to be 0.
This is possible when x=-1.
Therefore, at the vertex point x=h=-1 and y=k=9.

The following graph presents the graph of the function with its vertex point. 

vertex form graph 2

Solving a quadratic equation:

A quadratic equation is an equation that has a squared term (a variable multiplied by itself) and is usually written in a standard form ax2+bx+c=0 where a≠0 (x is a variable and a, b and c are constants). For example:

3x2+7x-1=0 is a quadratic equation with the parameters a=3, b=7 and c=-1.

-x2+7x=0 is a quadratic equation with the parameters a=-1, b=7 and c=0.

-5x2-1=0 is a quadratic equation with the parameters a=-5, b=0 and c=-1.

7x-1=0 is a not a quadratic equation since a=0, this equation is linear.

Solving a quadratic equation without x term (b=0)

Since we have only 2 numbers in the equation, we can isolate x2 and solve by taking a square root from each side of the equation.

Note that when we take square root, we get 2 answers, a positive answer and a negative answer. This is because a square of a negative number is positive. Therefore, don’t forget the negative answer! For example: √4x2=2x or -2x, since (2x)2=4x and also (-2x)2=4x.

Steps for solving a quadratic equation without x term (b=0):
1. Isolating x.
2. Taking a square root from each side of the equation.
3. Don’t forget that there is a positive answer and a negative answer.

Consider this example:

Solve 4x2-16=0.

4x2-16=0
4x2=16
4x2/4=16/4
x2=4
√x2=√4
x=2, x=-2

Note that x=-2 is also a solution for the equation.

Solving a quadratic equation without the constant term (c=0)

In these questions we need to solve a quadratic equation in its standard form that is ax2+bx (c=0). We can factor out x as a common factor getting x(ax+b)=0.
The zero product property states that if the multiplication of 2 numbers is equal to 0 then one of the numbers must be equal to 0, therefore x=0 or ax+b=0. We can solve ax+b=0 since it is a simple linear equation.

Consider the following example:

What are the solutions for the equation 2x2-4x=0?

2x2-4x=0
x(2x-4)=0
x=0 or 2x-4=0
2x=4
x=2

Checking the answer:
For x=0 solution: 2x2-4x=2*02-4*0=0
For x=2 solution: 2x2-4x=2*22-4*2=2*4-8=8-8=0

Solving a quadratic equation by factoring

We need to solve a quadratic equation in its standard form that is ax2+bx+c=0. Instead we can write it in its factored form (ax+c)(bx+d)=0.

The zero product property states that if the multiplication of 2 numbers is equal to 0 then one of the numbers must be equal to 0. In the factored formula equation (ax+c)(bx+d)=0 that means that ax+c= 0 or bx+d=0. Since a,b c and d are constants that we know we can find the x values from the factored formula by solving this 2 linear equations. For example: in the function y=(x+2)(x-4)=0, the intercepts with the x axis are at the points x=-2 and x=4.

Steps for solving a quadratic equation with factoring:
Step 1: Rewrite the standard form of the quadratic equation ax2+bx+c=0 as its factored form (ax+c)(bx+d)=0.
Step 2: Solve 2 linear equations by making each equation equal to 0: ax+c= 0 or bx+d=0. The solutions to the factored form equation are also the solutions to the standard form equation.

Consider the following example:

What are the solutions of the equation 2x2-6x-8=0?

Step 1: Rewriting the standard form of the quadratic equation to its factored form.
This was explained before.
First we can factor out 2 as a common factor getting 2(x2-3x-4)=0
We know that the factored form is (x+a)(x+b).
We know that ab=-4
We know that bx+ax=-3x, x(b+a)=-3x, b+a=-3.

The numbers which multiplication is equal to -4 and their sum is equal to -3 are -4 and 1, therefore the factored form is 2(x-4)(x+1).

Step 2: Solving the equations:
We will write the factored form as equal to zero 2(x-4)(x+1)=0. According to the zero product property we know that x-4=0 or x+1=0. Meaning that x=4 or x=-1 are the solutions for the equation.

Checking the answers:
2(4-4)(x+1)=2*0*(x-1)=0
2(x-4)(-1+1)=2*(x-4)*0=0

Solving a quadratic equation with the quadratic formula

Quadratic formula:
x=(-b±√(b2-4ac))/2a

For example: x2+x-30=0
a=1, b=1, c=-30

x=(-1±√(12-4*1*-30))/(2*1))
x1=(-1+11)/2=5
x2=(-1-11)/2=-6

For a standard form of a quadratic equation ax2+bx+c=0 the quadratic formula is
x=(-b±√(b2-4ac))/2a

Steps for solving a quadratic equation with the quadratic formula:
Step 1: Write the quadratic equation in its standard form ax2+bx+c=0.
Step 2: Plug the constants a, b and c into the quadratic formula x=(-b±√(b2-4ac))/2a.
Step 3: Evaluate x by solving the formula. Note that since we have a square root in the formula, we will have 2 solutions √(b2-4ac) and -√(b2-4ac).

Consider the following example:

What are the solutions to the equation x2+x=30?

Step 1: Writing the quadratic equation in its standard form
x2+x=30
x2+x-30=0
a=1, b=1, c=-30

Step 2: Plugging the constants a, b and c into the quadratic formula
x=(-b±√(b2-4ac))/2a
x=(-1±√(12-4*1*-30))/(2*1))
x=(-1+√(1+4*1*30))/2

Solution 1: (-1+√121)/2=(-1+11)/2=10/2=5
Solution 2: (-1-√121)/2=(-1-11)/2=-12/2=-6

Checking the answers:
x2+x-30=0
52+5-30=0, 0=0
(-6)2-6-30=0, 36-6-30=0, 0=0

Determining how many solutions a quadratic equation has

No real solution:
If b2-4ac<0 the quadratic equation has no real solution

For example: a=2, b=1, c=1
b2-4ac=1-4*2*1=-7<0 (no solution)
x=(-b±√-7)/2a (no solution)

1 real solution:
If b2-4ac=0 the quadratic equation has 1 real solution: 
x=-b/2a

For example: a=1, b=2, c=1
b2-4ac=4-4*1*1=0 (one solution)
x=(-b±√(0)/2a
x=(-2±√0)/2=-1

2 real solutions:
If b2-4ac>0 the quadratic equation has 2 real solutions:
x1=(-b+√(b2-4ac))/2a 
x2=(-b-√(b2-4ac))/2a

For example: a=1, b=-2, c=-8
b2-4ac=4-*1*-8=36>0 (two solutions)
x1=(-b+√(b2-4ac)/2a=(2+√36)/2=4
x2=(-b-√(b2-4ac)/2a=(2-√36)/2=-2

The quadratic equation can have 2 solutions, one solution or no solution. The number of the solutions depends on the value of the expression under the root b2-4ac. This expression is called discriminant

We know thar for a standard form of a quadratic equation ax2+bx+c=0 the quadratic formula is x=(-b±√(b2-4ac))/2a. When we evaluate x using this quadratic formula, we take a square root from the expression b2-4ac.

If the discriminant b2-4ac is negative we can’t take the root, because squared number is always positive. In this case the equation has no real solution.

If the discriminant b2-4ac is zero, the equation has one real solution that is x=-b/2a.

If the discriminant b2-4ac is positive, the equation has two real solutions x=(-b+√(b2-4ac))/2a and x=(-b√(b2-4ac))/2a.

Consider the following example:

What are the solutions of the equation x2=-2x-1?

Step 1: Writing the quadratic equation in its standard form:
x2=-2x-1
x2+2x+1=0
a=1, b=2, c=1

Step 2: Plugging the constants a, b and c into the quadratic formula:
x=(-b±√(b2-4ac))/2a
x=(-2±√(22-4*1*1))/(2*1)
x=(-2±√(4-4))/2
x=(-2±√0)/2
The discriminant is zero therefore we have 1 solution
x=-2/2=-1 the equation has only one solution x=-1.

Checking the answer:
x2=-2x-1
(-1)2=-2*-1-1
1=2-1
1=1

Consider the following example:

What are the solutions of the equation 2x2=-x-1?

Step 1: Writing the quadratic equation in its standard form:
2x2=-x-1
2x2+x+1=0
a=2, b=1, c=1

Step 2: Plugging the constants a, b and c into the quadratic formula:
x=(-b±√(b2-4ac))/2a
x=(-1±√(12-4*2*1))/(2*2)
x=(-1±√(1-8))/4

The discriminant is negative therefore the equatuon has no solution.
We can’t take a square root from -7 therefore the equation has no solution.

Consider the following example:

What are the solutions of the equation x2=2x+8?

Step 1: Writing the quadratic equation in its standard form:
x2=2x+8
x2-2x-8=0
a=1, b=-2, c=-8

Step 2: Plugging the constants a, b and c into the quadratic formula:
x=(-b±√(b2-4ac))/2a
x=(–2±√((-2)2-4*1*-8))/(2*1)
x=(2±√(4+32))/2
The discriminant is positive therefore we have 2 solutions
x=(2±√36)/2
x=(2±6)/2
We have two answers x1=8/2=4 or x2=-4/2=-2

Checking the answers:
First answer x1=4:
x2-2x-8= 42-2*4-8=16-8-8=0

Second answer x2=-2:
x2-2x-8= (-2)2-2*-2-8=4+4-8=0

Writing the factored form of a quadratic equation using the solutions from the quadratic formula

In order to write the factored form of the quadratic equation with the solutions we found using the quadratic formula we need to change the signs inside the brackets. If the solution is x=a in order for the expression (x+a) to be zero we need to write (x-a), that will give us a-a=0.

Consider the following example:

The equation x2-2x-8=0 has 2 solutions x=4 and x=-2. What is the factored form of the equation?

The factored form is (x-4)(x+2)=0, because plugging the solutions x=4 or x=-2 will give us zero as we need: (4-4)=0 or (2-2)=0.

We can also check the factored form by solving (x-4)(x+2).
(x-4)(x+2)=x2+2x-4x-8=x2-2x-8 we see that factored form is equal to the standard form.
The solutions for x2-2x-8=0 are x=4 or x=-2.
The solutions for (x-4)(x+2)=0 are x=4 or x=-2.

Quadratic word problems

There are 2 main types of word problems with exponential functions:  calculating the area of a rectangle problems and calculating height versus time problems. 

The skills required are:
Writing a quadratic function from a word problem.
Solving a given quadratic function according to a word problem.

Word problems- calculating the area of a rectangle

The formula of the area of a rectangle A=lw is its length l multiplied by its width w. If we receive data about the parameter l or w, we can express one parameter in terms of the other. Then we will have only one variable x and we will be able to write a factored form of a quadratic equation.

Consider the following example:

Rectangles length is 5 inches bigger than its width. What is the function representing its area?

w=x
l=x+5
A=x*(x+5)= x2+5x

Word problems- calculating height versus time

The relationship between height (h) and time (t) of falling objects is a quadratic function in its standard form h(t)=-at2+bt+c.
h(t)- the output- height (usually in feet)
t- the variable- time after the object is thrown (usually in seconds)
a coefficient that represents the object’s acceleration due to gravity (therefore a is negative)
b coefficient represents the initial velocity of the object
c- initial height of the object above the ground level (the height 0 is the ground, usually in feet)

Consider the following example:

The ball is launched up in the air 10 feet above the ground.
The function that models the height of the ball h, in feet, t seconds after being launched is h(t)=-t2+3t+10.

What is the height of the ball 0.05 minutes after it was launched?

We know that the time in the question is asked in minutes and the time in the formula is measured in seconds. 1 minute= 60 seconds, 0.05 minutes= 60*5/100 seconds= 3 seconds, therefore the time for the formula is t=3.
h(t)=-t2+3t+10
h(t=3)=-32+3*3+10=-9+9+10=10 feet.

Consider the following example:

The ball is launched up in the air 10 feet above the ground. The function that models the height of the ball h, in feet, t seconds after being launched is h(t)=-t2+3t+10. When will the ball be at the launching height?

The ball will go up and then fall down, the question is when will it be at the initial launching height? The launching height is 10, therefore we know that h(t)=-t2+3t+10=10.
-t2+3t+10=10
-t2+3t=0
t(-t+3)=0
t=0 or t=3

The answer t=0 is the launching point that we already know. The second answer t=3 is the answer. The ball will be again at the launching height of 10 feet after 3 seconds.

Consider the following example:

The ball is launched up in the air 10 feet above the ground. The function that models the height of the ball h, in feet, t seconds after being launched is h(t)=-t2+3t+10. How many seconds after launch will the ball hit the ground?

Since we are asked about the moment the ball will hit the ground, we know that the height h should be equal to 0.
h(t)=-t2+3t+10=0

We can solve with the quadratic formula x=(-b±√(b2-4ac))/2a
a=-1, b=3, c=10
x=(-3±√(32-4*-1*10))/(2*-1)
x=(-3±√(9+40))/-2
x=(-3±7)/-2
x=-10/-2=5 or x=4/-2=-2

Since we are looking for time the answer can’t be negative, therefore the only answer is t=5 seconds.

Checking the answer:
For x=5: -t2+3t+10=-52+3*5+10=-25+15+10=0

Finding the function of the parabola from its graph

We can find the standard form, the quadratic form and the vertex form equations using the vertex coordinates and the x and y intercepts coordinates from the graph.

Finding the factored form of the quadratic function from its graph

Recall that the factored form of the quadratic function is y=a(x-b)(x-c), it shows the x intercepts of the parabola so that the intercepts are (b,0) and (c,0).

Given the x intercepts coordinates from the graph, we know the values of b and c and we can write them in the function y=a(x-b)(x-c).  We are left with 3 variables: x, y (we need them for the function equation) and a. In order to find a we can plug any point into the equation, except the y intercepts values that we used.

Consider the following example:

Find the factored form of the graph below.

Finding the function of the parabola from its graph

Step 1: Identifying the vertex and the x intercept points:
The vertex is (-1,-8)
The x intercepts are (1,0) and (-3,0)

Step 2: Writing b and c values in the function y=a(x-b)(x-c):
In y=a(x-b)(x-c) we have b=1 and c=-3 so that the function is y=a(x-1)(x – – 3)=a(x-1)(x+3)
Checking: if x=1 we have y=a*0*4=0
Checking: if x=-3 we have y=a*-4*0=0

Step 3: Finding the value of a:
We know that the function is y=a(x-1)(x+3) and what is left is to find the value of a.
We can plug x and y values of any point from the graph.
We can plug the vertex coordinates (-1,-8) and get
-8=a(-1-1)(-1+3)
-8=-2*2*a
-8=-4a
a=2

The factored function is y=2(x-1)(x+3).

Finding the vertex form of the quadratic function from its graph

Given the vertex coordinates from the graph, we know the values of h and k and we can write them in the function y=a(x-h)2+k. We are left with 3 variables x, y (we need them for the function equation) and a. In order to find a we can plug any point into the equation, except the vertex values that we used.

Consider the following example:

Find the vertex form of the graph below.

Step 1: Identifying the vertex and the x intercept points:
The vertex is (-1,-8)
The x intercepts are (1,0) and (-3,0)

Step 2: Writing h and k values in the function y=a(x-h)2+k:
y=a(x-h)2+k
Since the vertex is (-1,-8) we know that k=-8 and h=-1.
y=a(x+1)2-8 (if we plug here x=-1 we get y=-8 and this is what we need for the vertex)

Step 3: Finding the value of a:
We know that the function is y=a(x+1)2-8 and what is left is to find the value of a.
We can plug x and y values of any point from the graph.
We can plug one of the x intercepts coordinates (1,0) and get
y=a(x+1)2-8
0= 4a-8
a=2

We can plug the other x intercept coordinates (-3,0) and get the same answer
y=a(x+1)2-8
0=4a-8
a=2

The vertex function is y=2(x+1)2-8

Finding the standard form of the quadratic function from its graph

We know the coordinates of 3 points from the graph: the vertex and the x intercepts. The standard form of a quadratic equation is y=ax2+bx+c. We can plug the (x,y) coordinates of the 3 point given in the graph and get 3 equations with 3 variables a, b and c. We can solve the equations and calculate the values of a, b and c.

Note that we can also find the standard form by finding the vertex form or the factored form and opening brackets.

Consider the following example:

Find the standard form of the graph below.

Finding the function of the parabola from its graph

Step 1: Identifying the vertex and the x intercept points:
The vertex is (-1,-8)
The x intercepts are (1,0) and (-3,0)

Step 2: writing 3 equations with 3 variables a, b and c:
y=ax2+bx+c
For (-1,-8) we get -8=a-b+c
For (1,0) we get 0=a+b+c
For (-3,0) we get 0=9a-3b+c

Step 3: solving the equations:
Isolating a from equation 1:
a=-8+b-c

Plugging a=-8+b-c into equation 2:
0=a+b+c
0=-8+b-c+b+c
2b=8
b=4
now we know that
a=-8+b-c=-8+4-c=-4-c

Plugging a=-4-c and b=4 into equation 3:
0=9a-3b+c
0=9(-4-c)-3*4+c
0=-36-9c-12+c
0=-48-8c
-8c=48
c=-6

Finding a by plugging b=4 and c=-6 into a=-8+b-c:
a=-8+4+6=2

The standard form is y=2x2+4x-6

Finding the standard form from the vertex form by opening brackets:

As we saw earlier the vertex form of the graph was is y=2(x+1)2-8

Opening brackets will get us
y=2(x+1)2-8
y=2(x2+2x+1)-8
y=2x2+4x+2-8

The standard form is y=2x2+4x-6

Finding the standard form from the factored form by opening brackets:

As we saw earlier the factored form of the graph was is y=2(x-1)(x+3)

Opening brackets will get us
y=2(x-1)(x+3)
y=2(x2+3x-x-3)
y=2(x2+2x-3)
y=2x2+4x-6

quadratic equations and functions formulas sheets

Exponential expressions on the SAT test

exponential expressions SAT image

Exponential expressions on the SAT test

SAT Subscore: Passport to Advanced Mathematics

Exponential expression includes 3 components: coefficient, base and exponent. For example 5x3.

The exponent is the number of times we multiply x (in the example the exponent is 3, since x3=x*x*x).
The base is the variable (in the example the base is x).
The coefficient is the number multiplied by the variable (in the example the coefficient is 5, since 5x3=5*x3.

On the SAT test exponential functions are part of Passport to Advanced Mathematics subscore questions. 

This topic contains questions that are solved using formulas. Let’s look at formulas list, different types of exponemtial functions questions and examples for each one of them.

what are the components of an exponential expression

In exponential functions questions you need to calculate the values of exponential expressions or solve equations with exponents and radicals.

The skills required in exponential functions questions are:
Dealing with different types of exponents: positive exponents, zero exponents and rational exponents (exponents with fractions) and dealing with radicals (roots). You also need to know how to solve word problems with exponential functions: population growth and decay problems and compounding interest problems. 

Performing operations with exponential expressions:
Changing the base of exponential expression.
Raising an exponential expression to an exponent.
Adding, subtracting multiplying and dividing exponential expressions.
Multiplying polynomial expressions with the FOIL formula.

Performing operations with radicals (roots):
Rewriting radicals as exponents.
Making calculations with exponents and radicals.
Multiplying and dividing radicals.

Solving word problems:
Writing an exponential function from a word problem.
Solving a given exponential function according to a word problem.
Rounding answers to the nearest whole number.
Calculating the net value of the change in the population.
Calculating the interest that is compounded between 2 periouds.
Changing the time units (changing the exponent of the expression).

What are the 4 types of exponents in the exponential expressions?

  1. Positive exponents are exponents with positive numbers. For example: x5.
  2. Zero exponents have 0 as the exponent and they all are equal to 1. For example: x0=1000=1.
  3. Negative exponents are exponents with negative numbers. For example: x-3.
  4. Rational exponents or radicals.
What are the 4 types of exponents in the exponential expressions?

What are Monomial and Polynomial expressions?

Monomial is a term of a form axn for constant a and none-negative integer n (the meaning of the word “mono” is one).
Note that the power can be zero so that the monomial will be equal to a constant a. Since x0=1 we know that ax0=a*1=a. For example, 8x0=8*1=8.

Consider the following examples:
5x7 is a monomial because 5 is constant and 7 is a non-negative integer.
4x is not a monomial because x is a variable (and not a non-negative integer).
X1/4 is not a monomial because 1/4 is a fraction (and not a non-negative integer).

Polynomial expression is composed of different monomials (the meaning of the word “poly” is many).
Binomial is a polynomial with 2 monomials, for example 5x7-5.
Trinomial is a polynomial with 3 monomials, for example 5x7+x4-5

Dealing with negative exponents and zero exponent:

Negative exponents formula:
x-n=1/xn

For example: x-3=1/x3

Zero exponent formula:
x0=1, x≠0

For example: 1000=1

The rule for negative exponents:

A positive base with a negative exponent is equal to 1 divided by the base raised to the opposite of the exponent. Meaning that in order to change the sign of the base from negative to positive, we must move the base between the numerator and the denominator. The formula for a negative exponent is x-n=1/xn
For example: x-3=1/x3

Consider this example:

x=a-n/4, n>0

What is a equal to in terms of x?
(a=?)

First, we will rewrite the expression making the negative exponent positive using the formula x-n=1/xn
x=a-n/4
x=1/an/4

Multiply both sides by an/4
an/4x=1*an/4/an/4
an/4x=1

Divide both sides by x
an/4=1/x

In order to cancel the exponent of a we raise both sides by 4/n
a=1/x4/n

Another way to solve the question is to raise both sides of the equation by the opposite exponent to -n/4 which is -4/n (since (-n/4)*(-4/n)=1).
x=a-n/4
x-4/n=a

We can also write this as
a=1/x4/n

 

The rule for an exponent that is equal to zero:

Every base that is raised to an exponent of 0 equals to 1. The formula for a zero exponent is x0=1

X0=1, 1000=0……

Note that the base itself can’t be 0 (x≠0).

Changing the base of an exponential expression:

We can write the same exponential expression with different bases. In order to find the equal expressions with different bases we need to check if we can make the base smaller or bigger.

1.Making the base smaller– Does the base have a square root or a cube root?
For example: 92=(32)2=34  (We changed the base from 9 to 3).
For example: 8=81=23 (We changed the base from 8 to 2).

2.Making the base bigger– Do we have an exponent in the expression?
For example: 92=811=81 (We changed the base from 9 to 81).

3.Making the base bigger– this is always possible. We can raise the base to an exponent and make a root to inverse the operation.
For example: 8=√64 or 641/2 (We changed the base from 8 to 64)

Consider this example:

If a2/b=16 for a and b that are positive integers, what are the values of a and b?

If we want the base on the left side of the equation to be 16 then we need to solve 162/b=161
2/b=1
b=2
a=16

Another option is to change the base of 16 to 4 by writing 16 as 42. Now we need to solve 4 2/b=42
2/b=2
b=1
a=4

Another option is to square the base of 16 to 256 by writing 16 as 2561/2. We need to solve 2562/b=2561/2
2/b=1/2
b=4
a=256

The rules for raising an exponential expression to an exponent:

In these questions we have an exponential expression and we need to raise it to another exponent. In order to simplify the expression, we need to open the brackets by raising everything that is in the brackets to the exponent.

Formula- the base is constant:

(ab) n=abn

For example: (34) 2=34*2=38

Formula- The base includes a variable:

(axm) n=anxmn

For example: (2x2) 3=23x2*3=8x6

When the base is a constant (no variable) the formula is (ab) n=abn
For example: (34) 2=32*4=38
We can solve without using the formula: (34) 2=(3*3*3*3)*(3*3*3*3)=38

When the base also includes a variable, the formula is (axm) n=anxmn with 3 steps:

  1. Raise the coefficient to the exponent
  2. Keep the base
  3. Multiply the exponents

For example: (2x2) 3=23x2*3=8x6
We can solve without using the formula: (2x2) 3=2x2*2x2*2x2=8*x*x*x*x*x*x=8x6

The rules for adding and subtracting exponential expressions:

We can add or subtract only exponential expressions that have the same base and the same exponent. This is done by combining like terms using the formula axn±bxn=(a±b)xn

Adding exponential expressions formula:

axn+bxn=(a+b)xn

For example: 3x3+2x3= (3+2)x3=5x3

Subtracting exponential expressions formula:

axn-bxn=(a-b)xn

For example: 3x3-2x3= (3-2)x3=x3

If we add or subtract expressions that have the same base and the same exponent, we need to use the formula axn±bxn=(a±b)xn with 3 steps:

  1. Keep the base.
  2. Keep the exponent.
  3. Add or subtract the coefficients.

Note that if the bases are different (3x3-2y3) or the exponents are different (3x5-2y3), we can’t add or subtract the expressions.

For example: 3x3-2x3=?
In both of the expressions the base is x and the exponent is 3 therefore we can use the formula.
3x3-2x3= (3-2)x3=x3
We can also solve the expression as 3x3-2x3= x3+x3+x3-x3-x3= x3

 

Adding and subtracting polynomials with different exponents and different bases:

If we add or subtract polynomials with different exponents and different bases we first need to group like terms and then follow the steps listed above.

  1. Group like terms
  2. Keep the bases
  3. Keep the exponents
  4. For each group off equal bases and equal exponents add or subtract the coefficients.

Consider this example:

Solve 8x3+2x-5x3-(5-4x)

Open brackets:
8x3+2x-5x3-5- -4x=8x3+2x-5x3-5+4x

Group like terms (find equal bases and equal exponents):
8x3-5x3+4x+2x-5

For each group off equal bases and equal exponents add or subtract the coefficients with the formula axn±bxn=(a±b)xn:
(8-5)x3+(2+4)x-5
group 1 includes expressions with x3, group 2 includes expressions with x1 and group 3 includes expressions with x0

Continue solving: 3x3+6x-5

The rules for multiplying and dividing exponential expressions:

We can multiply or divide only exponential expressions that have the same base or the same exponent.

Multiplying and dividing exponential expressions with the same base:

Formula for myltiplying exponential expressions with same base:
axm*bxn=(a*b)xm+n

For example: 2x2 *4x3=8x5

Formula for dividing exponential expressions with the same base:
axm/bxn=a/b*xm-n

For example: 6x5 /2x3=3x2

If we multiply expressions that have the same base, we need to use the formula axm*bxn=(a*b)xm+n with 3 steps:

  1. Multiply the coefficients
  2. Keep the base
  3. Add the exponents

For example: 2x2 *4x3=(2*4)x2+3 =8x5

We can also solve the expression without using the formula:  2x2 *4x3=2*4*x*x*x*x*x=8*x5=8x5

If we divide expressions that have the same base, we need to use the formula axm/bxn =a/b*xm-n with 3 steps:

  1. Divide the coefficients
  2. Keep the base
  3. Subtract the exponents

For example: 6x5 /2x3=(6/2)x5-3 =3x2

We can also solve the expression without using the formula:  6x5 /2x3=6/2*(x*x*x*x*x )/(x*x*x)=3*x*x =3x2

Multiplying exponential expressions with the same exponents:

Formula for myltiplying exponential expressions with the same exponents: 

axn*byn=ab(xy)n

For example: 6x3 *2y3=12(xy)3

Formula for dividing exponential expressions with the same exponents:

(axn)/(byn)=(a/b)(x/y)n

For example: 6x3 /2y3=3(x/y)3

If we multiply expressions that have the same exponent, we need to use the formula axn*byn=ab(xy)n with 3 steps:

  1. Multiply the coefficients
  2. Multiply the bases
  3. Keep the exponent

For example: 6x3 *2y3=6*2*x3*y3 =12(xy)3

If we divide expressions that have the same exponent, we need to use the formula (axn)/(byn)=(a/b)(x/y)n with 3 steps:

  1. Divide the coefficients
  2. Divide the bases
  3. Keep the exponent

For example: 6x3 /2y3=(6/2)*(x3/y3)=3(x/y)3

Note that if the bases are different and the exponents are different (3x5*2y3 or 4x5/2y3), we can’t multiply or divide the expressions. The only thing we can do is to multiply or divide the coefficients.

Note that if the bases and the exponents are the same, we can solve the expression with each one of the formulas.
For example:
2x3*4x3=2*4*x3+3=8x
2x3*4x3=2*4*(x*x) 3 =8*(x2) 3 =8x6

The rules for multiplying polynomial expressions:

FOIL Formula for multiplying two binomials:
(ax+b)(cx+d)=ax*cx+ax*d+b*cx+b*d

For example: (2x+5)(3x-3)=2x*3x+2x*-3+5*3x+5*-3

Monomial is a term of a form axn for constant a and none-negative integer n. For example: 5x7.

Polynomial expression is composed of different monomials. For example: 5x7+x4-5.

Multiplying monomial and a polynomial:

Multiplying monomial  and a polynomial requires us to write the polynomial in brackets and then distribute by opening brackets. Then we need to continue solving the distributed terms according to the exponent rules explained above.

Assume that we need to multiply a monomial and a trinomial, for example multiply 3 and 5x6-4x-2.  We write the multiplication as 3(5x6-4x-2) and then solve by multiplying 3 by every number inside the brackets.

3(5x6-4x-2)=3*5x6+3*-4x+3*-2=15x6-12x-6

Multiplying two binomials with FOIL method: