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 7 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. f(x)=(x-2)^2(x+1) 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 f(x)=0. f(x)=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 - 1, and 2.

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-2)^2(x+1) f(x)=(x-2)^2(x+1)
- 0.5 ( - 0.5-2)^2( - 0.5+1) 3.125
0 ( 0-2)^2( 0+1) 4
1 ( 1-2)^2( 1+1) 2
3 ( 3-2)^2( 3+1) 4
The points ( - 0.5, 3.125), ( 0, 4), ( 1, 2), and ( 3, 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.
f(x)=(x-2)^2(x+1)
f(x)=(x^2-2x(2)+2^2)(x+1)
f(x)=(x^2-4x+4)(x+1)
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Simplify right-hand side
f(x)=x^2(x+1)-4x(x+1)+4(x+1)
f(x)=x^3+x^2-4x(x+1)+4(x+1)
f(x)=x^3+x^2-4x^2-4x+4(x+1)
f(x)=x^3+x^2-4x^2-4x+4x+4
f(x)=x^3-3x^2+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.