Question

What is the equation of the following graph in vertex form?

Answer 1

$\bf \boxed{y=(x-{{ h}})^2+{{ k}}}\\\\ x=(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -----------------------------\\\\ \textit{vertex is at 4,-4}\implies \begin{cases} h=4\\ k=-4 \end{cases}\implies y=(x-4)^2+(-4) \\\\\\ \boxed{y=(x-4)^2-4}$

Answer 2

For
y=a(x-h)²+k
vertex is (h,k)

so

the vertex is (4,-4)
so
y=a(x-4)²-4
find a
we see the point
(0,12)
sub 0 for x and 12 for y and solve for a
12=a(0-4)²-4
12=a(-4)²-4
12=16a-4
add 4 both sides
16=16a
1=a


equation is
y=1(x-4)²-4 or
y=(x-4)²-4

A is answer