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.

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 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 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).

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).