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 = (x3)^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 = (y3)^2 + 9
x  9 = (y3)^2
x  9 = (y+3)^2
sqrt(x9) = y + 3
y = sqrt(x9)  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??
I need to find the inverse of y = 6x  x^2, given that x>/=3.
I simplified and got y = (x3)^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 = (y3)^2 + 9
x  9 = (y3)^2
x  9 = (y+3)^2
sqrt(x9) = y + 3
y = sqrt(x9)  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…
♣♦
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…
♣♦