the function has vertex at (-2, 8) and goes through point (0,4) (y-intercept)
since vertex form for parabola at (h,k) is
f(x) = a(x-h)² +k
the vertex form looks like: f(x) = a(x- (-2) )² + 8 = a(x+2)² + 8
we do not know a but we can find it with the point (0,4) that is on the graph
point (0,4) means when x = 0, f(x) = 4
since f(x) = a(x+2)² + 8, substitute f(x) with 4 and x with 0 and solve for a
4 = a(0+2)² + 8
4 = a(2)² + 8
4 = 4a + 8
-4 = 4a
a = -1
so it is f(x) = -(x+2)² + 8 but we need to expand(x+2)² because all choices r in standard form
(x+2)² = (x+2)(x+2) = x(x+2) + 2(x+2) = x² + 2x + 2x + 4 = x² + 4x + 4 so
f(x) = -(x+2)² + 8
= -(x² + 4x + 4) + 8
= -x² -4x - 4 + 8
= -x² - 4x + 4
only choice A fits this