please help!!! need this done asap

Question

Determine the standard form of a quadratic function whose graph has a y intercept of 8 and a vertex at (-2,-4)

(the answer is f(x)= 3x^2+12x+8, I've been given the answer but i need to show the work

Answer 1

The formula has the form y = a(x-h)² + k, where (h,k) is the vertex.

So, we have y = a(x + 2)² - 4.

Since (0,8) is on the graph, 8 = 4a - 4

4a = 12

So, a = 3

y = 3(x + 2)² - 4

y = 3(x² + 4x + 4) - 4

y = 3x² + 12x + 8