Posted by Diana on Friday, July 15, 2011 at 4:21pm.
first one:
y = 4/(x-4)^2
inverse is
x = 4/(y-4)^2
solving this for y .....
(y-4)^2 = 4/x
y - 4 = ± 2/√x
y = 4 ± 2/√x
2nd .....
inverse is
x = y^2 - 8y + 16
y^2 - 8y + 16 - x = 0
use the quadratic equation
where a=1, b=-8 and c = 16-x to solve for y
thanks i got the first one now!
for the second one I used the quadratic eqn and got 4 but that is not the answer?
y = (8 ± √(64-4(1)(16-x))/2
= (8 ± √(4x) )/2
= (8 ± 2√x)/2
= 4 ± √x
check: let x = 3 into the first
y = 9 - 24 + 16 = 1
put x = 1 into the second:
y = 4 ± √1 = 5 or 3
notice the the inverse of each of the equations is not a function.
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