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 1 Page 212

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)=1/2(x+1)(x-4)^2 Let's start by finding its zeros.

Zeros of the Function

To do so, we need to find the values of x for which f(x)=0. f(x)=0 ⇔ 1/2(x+1)(x-4)^2=0 Since the function is already written in factored form, we will use the Zero Product Property. Notice, we do not set the coefficient 12 equal to zero as it produces a false statement.
(x+1)(x-4)^2=0
lcx+1=0 & (I) (x-4)^2=0 & (II)
lx=- 1 (x-4)^2=0
lx=- 1 x-4=0
lx=- 1 x=4
We found that the zeros of the function are - 1 and 4.

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 12(x+1)(x-4)^2 f(x)= 12(x+1)(x-4)^2
- 12 12(( - 12)+1)(( - 12)-4)^2 5.063
0 1/2( 0+1)( 0-4)^2 8
2 1/2( 2+1)( 2-4)^2 6
3 1/2( 3+1)( 3-4)^2 2
The points ( 12, 5.063), ( 0, 8), ( 2, 6), and ( 3, 2) 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)=1/2(x+1)(x-4)^2
f(x)=1/2(x+1)(x-4)(x-4)
â–Ľ
Distribute (x-4) & Simplify
f(x)=1/2(x+1)(x(x-4)-4(x-4))
f(x)=1/2(x+1)(x^2-4x-4(x-4))
f(x)=1/2(x+1)(x^2-4x-4x+16)
f(x)=1/2(x+1)(x^2-8x+16)
â–Ľ
Distribute (x+1) & Simplify
f(x)=1/2(x^2(x+1)-8x(x+1)+16(x+1))
f(x)=1/2(x^3+x^2-8x(x+1)+16(x+1))
f(x)=1/2(x^3+x^2-8x^2-8x+16(x+1))
f(x)=1/2(x^3+x^2-8x^2-8x+16x+16)
f(x)=1/2(x^3-7x^2+8x+16)
â–Ľ
Distribute 1/2 & Simplify
f(x)=1/2(x^3)+1/2(-7x^2)+1/2(8x)+1/2(16)
f(x)=1/2x^3-7/2x^2+4x+8
We can see now that the leading coefficient is 12, 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!