This is the question:
Find the quadratic function that has vertex (0, -9) and whose graph goes through the point (12, -729).
f (x) = _______________
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How can I solve this? Please me.
2007-03-30 11:04:47 · 6 answers · asked by Jayla K 1 in Science & Mathematics ➔ Mathematics
y=mx+b
(-729-(-9))
----------- = M === -60
12-0
y=-60x-9
make it x^2 by dividing 60 into -5 and 12. 12 is canceled by the 12 in the problem
y=-5x^2-9
2007-03-30 11:10:22 · answer #1 · answered by AZKAOZ 2 · 0⤊ 1⤋
The vertex form for a parabola with vertex (h,k) is
y = a(x-h)^2 + k, so so far you know it's y = a(x-0)^2 + -9
or y = ax^2 - 9
Now plug in 12 for x and -729 for y to solve for a
-729 = 144a - 9
-720 = 144a
a = -5
y = -5x^2 - 9
2007-03-30 11:10:13 · answer #2 · answered by hayharbr 7 · 0⤊ 0⤋
If a parabola has vertex (p,q)
its equation is
y=a(x-p)^2+q in this case
y=a(x^2)-9
As it goes through (12,-729)
-729 =144a -9 so a= -720/144 =-5
y=-5*x^2-9
2007-03-30 11:16:45 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
y = a (x-h)^2 + k
plug the points in
-729 = a(12 - 0)^2 - 9
-720 = 144a
a = -5
so we know a = 5
y = -5(x-0)^2 - 9
y = -5x^2 - 9
2007-03-30 11:10:09 · answer #4 · answered by 7 · 0⤊ 0⤋
x^2 =2p(y+9)
12^2 =2p(-720)
2p= 144/-720
x^2= -.2(y+9)
-.2y -1.8 = x^2
y = -5x^2 -9
2007-03-30 11:28:17 · answer #5 · answered by ironduke8159 7 · 0⤊ 0⤋
6-4x42=yz(x)f
97862 =y,(y) and f
O.k.?
2007-03-30 11:08:42 · answer #6 · answered by Daisy-flower 3 · 0⤊ 1⤋