Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
8. Analyzing Graphs of Polynomial Functions
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Exercise 4 Page 216

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

C

Practice makes perfect
We want to find the zeros of the given polynomial function. h(x)=(x+2)^2(x+1)

Zeros of the Function

To do so, we need to find the values of x for which f(x)=0. y=0 ⇔ (x+2)^2(x+1)=0 Since the function is already written in factored form, we will use the Zero Product Property.
(x+2)^2(x+1)=0
lc(x+2)^2=0 & (I) x+1=0 & (II)
lx+2=0 x+1=0
lx=- 2 x+1=0
lx=- 2 x=- 1
We found that the zeros of the function are - 2, and - 1. We can stop here and recognize that the only graph with those zeros is graph C. We can also continue by graphing the function with a table of values as shown below.

Extra

Graphing the function using a table of values

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+2)^2(x+1) h(x)=(x+2)^2(x+1)
- 3 ( - 3+2)^2( - 3+1) - 2
- 1.5 ( - 1.5+2)^2( - 1.5+1) - 0.125
- 0.5 ( - 0.5+2)^2( - 0.5+1) 1.125
0 ( 0+2)^2( 0+1) 4
The points ( - 3, - 2), ( - 1.5, - 0.125), ( - 0.5, 1.125), and ( 0, 4) 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+2)^2(x+1)
â–Ľ
Rewrite (x+2)^2 as (x+2)(x+2)
h(x)=(x+2)(x+2)(x+1)
h(x)=(x(x+2)+2(x+2))(x+1)
h(x)=(x^2+2x+2(x+2))(x+1)
h(x)=(x^2+2x+2x+4)(x+1)
h(x)=(x^2+4x+4)(x+1)
â–Ľ
Distribute x+1
h(x)=x^2(x+1)+4x(x+1)+4(x+1)
h(x)=x^3+x^2+4x(x+1)+4(x+1)
h(x)=x^3+x^2+4x^2+4x+4(x+1)
h(x)=x^3+x^2+4x^2+4x+4x+4
h(x)=x^3+5x^2+8x+4
We can see now that the leading coefficient is 1, which is a positive number. Also, the degree is 3, which is an odd number. Therefore, the end behavior is down and up. Now, let's draw the graph!