Question

The function f(x) = x2 + 22x + 58 is translated 4 units to the right and 16 units up. What is the vertex form of the new function?

Answer 1

The answer to this equation is C.
C. (x + 7)2 – 47

Answer 2

Answer:

The new function is $f(x)=(x+7)^2-47$.

Step-by-step explanation:

The given function is

$f(x)=x^2+22x+58$

$f(x)=(x^2+22x)+58$

To find the vertex from add and subtract $(-\frac{b}{2a})^2$ in the parenthesis.

$(-\frac{b}{2a})^2=(-\frac{22}{2(1)})^2=11^2$

$f(x)=(x^2+22x+11^2-11^2)+58$

$f(x)=(x^2+22x+11^2)-121+58$

$f(x)=(x+11)^2-63$              $[\because (a+b)^2=a^2+2ab+b^2]$

The vertex of the given function is (-11,-63).

It is given that the function f(x) is translated 4 units to the right and 16 units up.

$(x,y)\rightarrow (x+4,y+16)$

$(-11,-63)\rightarrow (-11+4,-63+16)=(-7,-47)$

The vertex of new function is (-7.-47). So the new function is

$f(x)=(x+7)^2-47$