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