Question

Help with inverse equation
h(x)= 2 sqrt. (x+3)
x= 2 sqrt. (y+3)
x^2= 2 sqrt. (y+3)^2
x^2= 2y+6
-6. -6
x^2 -6= 2y
÷2. ÷2
(x^2 -6)/2= y
??????
The answer should be 1/4 x-3
I cannot get the 1/4 help please!

Answer 1

You need to get rid of the radical sign \(\sqrt{y+3}\), so you must square it ( \((\sqrt{y+3})^2\)) to get \(y+3\). But of course, you must square it on both sides! So in the left side, you'll get \(\frac{ x^2 }{ 4 }\) or \(\frac{ 1 }{ 4 }x^2\).

Answer 2

Once you've done that, you can now easily solve for y.