Question

the function $f(x) = x^2 + 22x + 58$ is translated 4 units to the right and 16 units up. what is the vertex form of the new function?
\bigcirc $(x - 11)^2 + 58$
\bigcirc $(x + 22)^2 - 121$
\bigcirc $(x + 7)^2 - 47$
\bigcirc $(x - 15)^2 + 94$

Explanation:

Step1: Convert to vertex form

Complete the square for \(f(x)=x^2 + 22x+58\).
The formula for completing the square is \(x^2+bx=(x + \frac{b}{2})^2-\frac{b^2}{4}\). Here, \(b = 22\), so \(\frac{b}{2}=11\), \(\frac{b^2}{4}=\frac{22^2}{4}=121\).
\(f(x)=(x + 11)^2-121 + 58=(x + 11)^2-63\). The vertex of \(f(x)\) is \((- 11,-63)\).

Step2: Apply translations

Translate 4 units right: \(x\)-coordinate of vertex becomes \(-11 + 4=-7\).
Translate 16 units up: \(y\)-coordinate of vertex becomes \(-63+16 = - 47\).

Step3: Write new vertex form

The vertex form of a parabola is \(y=(x - h)^2+k\), where \((h,k)\) is the vertex. Here, \(h=-7\), \(k = - 47\), so the equation is \(y=(x+7)^2-47\).

Answer:

\((x + 7)^2-47\)