[From: ] [author: ] [Date: 11-05-02] [Hit: ]
...I need to find the inverse of y = 6x - x^2, given that x>/=3.I simplified and got y = -(x-3)^2 + 9,......
322. This is part of a more complicated problem I keep getting stuck on....
I need to find the inverse of y = 6x - x^2, given that x>/=3.
I simplified and got y = -(x-3)^2 + 9, given that x >/= 3.
I graphed that parabola with vertex (3,9) and x intercept 6. I'm assuming I did all of that right... if not, then uh, help!

Anyways, this is what I have for finding the inverse:

x = -(y-3)^2 + 9
x - 9 = -(-y-3)^2
x - 9 = (y+3)^2
sqrt(x-9) = y + 3
y = sqrt(x-9) - 3

But, when I graph that, it clearly does not look like the inverse.
I think the inverse would have a y intercept at 6 since the original had a x intercept at 6, right??

-
y = -(x - 3)^2 + 9, for x ≥ 3
y - 9 = -(x - 3)^2
9 - y = (x - 3)^2 ← you only divided one side by -1
±√(9 - y) = x - 3 ← carry the ± until the last step
3 ± √(9 - y) = x
and since you started with only the right side,
you should end up with only the upper side in the inverse:
3 + √(9 - y) = x
so inverse is:
y = 3 + √(9 - x)

see
http://s584.photobucket.com/albums/ss282…

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