Inverse of Quadratics with Horizontal Transformations This might sound like a beginner question for a lot of the community, I apologize, but, I really need help understanding it. I was trying to find the inverse of the quadratic $ f(x)= (\frac 12x +2)^2 +4$. Using WolframAlpha, I was able to find that the inverse function was $ f(x) = ±  2 \sqrt{x - 4}-4$
My steps are the following;
$y=(\frac 12x +2)^2 +4$
$x=(\frac 12y +2)^2 +4$
$x-4=(\frac 12y +2)^2$
$ \sqrt{x-4} =\frac 12y +2$
$ \sqrt{x-4} -2 =\frac 12y$
$ 2\sqrt{x-4} -2 =y$
I don't understand why I am getting a different inverse equation than the original.

$y=(\frac 12x +2)^2 +4$
$x=(\frac 12y +2)^2 +4$
$x-4=(\frac 12y +2)^2$
Everything is right up to here.
$\sqrt{x^2} = |x|$, so you need a $\pm$ in front of the square root, because you need to account for both positive and negative values of $y$ for a given $x$ while solving for the inverse.
$\pm \sqrt{x-4} =\frac 12y +2$
$\pm \sqrt{x-4} -2 =\frac 12y$
You need to multiply both sides by $2$ (above, you only multiplied $\sqrt{x-4}$ by $2$, but not the $-2$).
Final answer: $y = \pm 2\sqrt{x-4} - 4 $
