Question

writing a quadratic function with a ≠ 1 in vertex form
what is $h(x) = -3x^2 - 6x + 5$ written in vertex form?
○ $h(x) = -3(x + 1)^2 + 2$
○ $h(x) = -3(x + 1)^2 + 8$
○ $h(x) = -3(x - 3)^2 - 4$
○ $h(x) = -3(x - 3)^2 + 32$

Explanation:

Step1: Factor out the coefficient of \(x^2\) from the first two terms

Given \(h(x)=-3x^{2}-6x + 5\), factor out \(-3\) from the first two terms:
\(h(x)=-3(x^{2}+2x)+5\)

Step2: Complete the square inside the parentheses

To complete the square for \(x^{2}+2x\), we take half of the coefficient of \(x\) (which is \(2\)), square it (\((\frac{2}{2})^{2}=1\)), and add and subtract it inside the parentheses. But since there is a factor of \(-3\) outside, we need to be careful.
\(h(x)=-3(x^{2}+2x + 1-1)+5\)
Rewrite \(x^{2}+2x + 1\) as \((x + 1)^{2}\):
\(h(x)=-3((x + 1)^{2}-1)+5\)

Step3: Distribute the \(-3\) and simplify

Distribute the \(-3\) into the parentheses:
\(h(x)=-3(x + 1)^{2}+3 + 5\)
Combine the constant terms:
\(h(x)=-3(x + 1)^{2}+8\)

Answer:

\(h(x)=-3(x + 1)^{2}+8\) (corresponding to the option: \(h(x) = -3(x + 1)^2 + 8\))