Please help I can't figure out these two.

Express f(x) in the form a(x − h)2 + k.
f(x) = −4x2 + 24x − 9

Find the standard equation of a parabola that has a vertical axis and satisfies the given conditions.

vertex (0, −7), passing through (3, 38)

Thank You!

5 answers

for the first, complete the square:

f(x) = -4(x^2 - 6x ......) - 9
= -4( x^2 - 6x + 9 -9) - 9
= -4( (x-3)^2 - 9) - 9
= -4(x-3)^2 + 36 - 9
= -4(x-3)^2 + 27

for the second, you know the vertex is (0,-7)
so f(x) = a(x-0)^2 - 7
= ax^2 - 7
but (3,38) lies on it, so
38 = a(9)
a = 38/9

f(x) = (38/9)x^2 -7
f(x) = -4(x^2-6x) - 9
= -4(x^2-6x+9) -4(-9) - 9
= -4(x-3)^2 + 27

f(x) = a(x-0)^2 - 7
38 = 9a-7
a = 5
f(x) = 5x^2 - 7
don't know where my - 7 went ????
Thank You Guys!!!
F(x) = -4x^2+24x-9.

h = -B/2A = -24/-8 = 3.
k = -4*3^2 + 24*3 - 9=-36 + 72 - 9 = 27.

F(x) = a(x-h)^2 + k.
F(x) = -4(x-3)^2 + 27.