Posted by **Eric** on Friday, March 22, 2013 at 5:02pm.

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!

- Algebra -
**Reiny**, Friday, March 22, 2013 at 5:11pm
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

- Algebra -
**Steve**, Friday, March 22, 2013 at 5:11pm
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

- go with Steve's end # Algebra -
**Reiny**, Friday, March 22, 2013 at 5:13pm
don't know where my - 7 went ????

- Algebra -
**Eric**, Friday, March 22, 2013 at 5:20pm
Thank You Guys!!!

- Algebra -
**Henry**, Saturday, December 5, 2015 at 10:06pm
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.

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