Question

1. convert each function into vertex form. (complete 2/3)

a. $f(x)=x^{2}+10x - 56$
b. $y = x^{2}+4x - 12$
c. $h(x)=4(x - 6)(x + 4)$

Explanation:

Part a: \( f(x) = x^2 + 10x - 56 \)

Step1: Recall vertex form formula

The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. For a quadratic in the form \( f(x)=x^2+bx + c \), we complete the square.

Step2: Complete the square for \( x^2 + 10x \)

Take the coefficient of \( x \), which is \( 10 \). Divide it by \( 2 \) to get \( \frac{10}{2}=5 \), then square it: \( 5^2 = 25 \).
We add and subtract this value inside the function:
\( f(x)=(x^2 + 10x + 25)-25 - 56 \)

Step3: Rewrite as a perfect square

The expression \( x^2 + 10x + 25 \) is a perfect square trinomial, which factors to \( (x + 5)^2 \).
So, \( f(x)=(x + 5)^2-25 - 56 \)

Step4: Simplify the constant terms

Calculate \( -25 - 56=-81 \).
So, \( f(x)=(x + 5)^2-81 \)

Part b: \( y = x^2 + 4x - 12 \)

Step1: Recall vertex form formula

Vertex form is \( y = a(x - h)^2 + k \). For \( y=x^2+bx + c \), complete the square.

Step2: Complete the square for \( x^2 + 4x \)

Coefficient of \( x \) is \( 4 \). Divide by \( 2 \): \( \frac{4}{2}=2 \), square it: \( 2^2 = 4 \).
Add and subtract \( 4 \) inside the function:
\( y=(x^2 + 4x + 4)-4 - 12 \)

Step3: Rewrite as a perfect square

\( x^2 + 4x + 4=(x + 2)^2 \), so \( y=(x + 2)^2-4 - 12 \)

Step4: Simplify the constant terms

\( -4 - 12=-16 \), so \( y=(x + 2)^2-16 \)

Part c: \( h(x) = 4(x - 6)(x + 4) \)

Step1: Expand the product

First, expand \( (x - 6)(x + 4) \) using the distributive property (FOIL method):
\( (x - 6)(x + 4)=x^2+4x-6x - 24=x^2 - 2x - 24 \)
Then multiply by \( 4 \): \( h(x)=4(x^2 - 2x - 24)=4x^2-8x - 96 \)

Step2: Factor out the coefficient of \( x^2 \) from the first two terms

Factor out \( 4 \) from \( 4x^2-8x \): \( h(x)=4(x^2 - 2x)-96 \)

Step3: Complete the square inside the parentheses

For \( x^2 - 2x \), coefficient of \( x \) is \( -2 \). Divide by \( 2 \): \( \frac{-2}{2}=-1 \), square it: \( (-1)^2 = 1 \).
Add and subtract \( 1 \) inside the parentheses:
\( h(x)=4((x^2 - 2x + 1)-1)-96 \)

Step4: Rewrite as a perfect square and simplify

\( x^2 - 2x + 1=(x - 1)^2 \), so:
\( h(x)=4((x - 1)^2-1)-96=4(x - 1)^2-4 - 96 \)

Step5: Simplify the constant terms

\( -4 - 96=-100 \), so \( h(x)=4(x - 1)^2-100 \)

Answer:

s:
a. \( \boldsymbol{f(x)=(x + 5)^2-81} \)
b. \( \boldsymbol{y=(x + 2)^2-16} \)
c. \( \boldsymbol{h(x)=4(x - 1)^2-100} \)