Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
8. Analyzing Graphs of Polynomial Functions
Continue to next subchapter

Exercise 9 Page 216

Use the Zero Product Property to find the zeros of the polynomial function.

Practice makes perfect

We want to graph the given polynomial function. h(x)=(x+1)^2(x-1)(x-3) Let's start by finding the zeros of the function.

Zeros of the Function

To do so, we need to find the values of x for which h(x)=0. h(x)=0 ⇔ (x+1)^2(x-1)(x-3)=0 Since the function is already written in factored form, we will use the Zero Product Property.
(x+1)^2(x-1)(x-3)=0
lc(x+1)^2=0 & (I) x-1=0 & (II) x-3=0 & (III)
lx+1=0 x-1=0 x-3=0
lx=-1 x-1=0 x-3=0
lx=-1 x=1 x-3=0
lx=-1 x=1 x=3
We found that the zeros of the function are - 1, 1, and 3.

Graph

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

x (x+1)^2(x-1)(x-3) h(x)=(x+1)^2(x-1)(x-3)
- 1.5 ( - 1.5+1)^2( - 1.5-1)( - 1.5-3) 2.813
-0.5 ( - 0.5+1)^2( - 0.5-1)( - 0.5-3) 1.313
0 ( 0+1)^2( 0-1)( 0-3) 3
2 ( 2+1)^2( 2-1)( 2-3) -9
The points ( - 1.5, 2.813), ( -0.5, 1.313), ( 0, 3), and ( 2, -9) are on the graph of the function. Finally, let's apply the Distributive Property. This will simplify the equation and determine the leading coefficient and degree of the polynomial function.
h(x)=(x+1)^2(x-1)(x-3)
h(x)=(x^2+2x+1)^2(x-1)(x-3)
â–Ľ
Simplify right-hand side
h(x)=(x^2(x-1)+2x(x-1)+1(x-1))(x-3)
h(x)=(x^3-x^2+2x(x-1)+1(x-1))(x-3)
h(x)=(x^3-x^2+2x^2-2x+1(x-1))(x-3)
h(x)=(x^3-x^2+2x^2-2x+x-1)(x-3)
h(x)=(x^3+x^2-x-1)(x-3)
h(x)=x^3(x-3)+x^2(x-3)-x(x-3)-1(x-3)
h(x)=x^4-3x^3+x^2(x-3)-x(x-3)-1(x-3)
h(x)=x^4-3x^3+x^3-3x^2-x(x-3)-1(x-3)
h(x)=x^4-3x^3+x^3-3x^2-x^2+3x-1(x-3)
h(x)=x^4-3x^3+x^3-3x^2-x^2+3x-x+3
h(x)=x^4-2x^3-4x^2+2x+3
We can see now that the leading coefficient is 1, which is a positive number. Also, the degree is 4, which is an even number. Therefore, the end behavior is up and up. Now, let's draw the graph!