Question

using completing the square 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?
$(x + 22)^2 - 121$
$(x - 11)^2 + 58$
$(x - 15)^2 + 94$
$(x + 7)^2 - 47$

Explanation:

Step1: Complete the square for $f(x)$

For $f(x)=x^2+22x+58$, take half of 22: $\frac{22}{2}=11$, square it: $11^2=121$. Rewrite the function:

$$\begin{align*} f(x)&=x^2+22x+121-121+58\\ &=(x+11)^2-63 \end{align*}$$

Step2: Apply horizontal translation

Translating 4 units right replaces $x$ with $x-4$:

$$ f(x-4)=(x-4+11)^2-63=(x+7)^2-63 $$

Step3: Apply vertical translation

Translating 16 units up adds 16 to the function:

$$ (x+7)^2-63+16=(x+7)^2-47 $$

Answer:

$(x + 7)^2 - 47$