Question

completing the square to write in vertex form
what is $f(x) = x^2 - 8x + 11$ written in vertex form?
\bigcirc $f(x) = (x - 4)^2 - 5$
\bigcirc $f(x) = (x - 4)^2 + 5$
\bigcirc $f(x) = (x + 4)^2 - 27$
\bigcirc $f(x) = (x + 4)^2 + 27$

Explanation:

Step1: Recall vertex form and completing the square

The vertex form of a quadratic function is \( f(x) = (x - h)^2 + k \), where \((h, k)\) is the vertex. To complete the square for \( f(x) = x^2 - 8x + 11 \), we take the coefficient of \( x \), which is \(-8\), divide by 2: \(\frac{-8}{2} = -4\), and square it: \((-4)^2 = 16\).

Step2: Rewrite the function

We rewrite the function by adding and subtracting 16:
\( f(x) = x^2 - 8x + 16 - 16 + 11 \).

Step3: Form the perfect square trinomial

The first three terms \( x^2 - 8x + 16 \) form a perfect square trinomial, which factors as \( (x - 4)^2 \). Then we simplify the constants: \( -16 + 11 = -5 \). So \( f(x) = (x - 4)^2 - 5 \).

Answer:

A. \( f(x) = (x - 4)^2 - 5 \)