Question 638927
In {{{y = a(x-h)^2 + k}}}, the vertex is (h,k).
The vertex is (2, -4) therefore h = 2 and k = -4.
{{{y = a(x-2)^2 - 4}}}, substitute (1,-2) and solve for "a".
{{{-2 = a(1 - 2)^2 - 4}}}
{{{-2 = a - 4}}}
{{{a = 2}}}
{{{y = 2(x-2)^2 - 4}}}.
Answer: {{{highlight(f(x) = 2x^2 - 8x + 4)}}}