X and Y intercept basic issue Finding the Intercepts of the Graph of an Equation
Given an equation involving x and y, we ﬁnd the intercepts of the graph as follows


*

*x-intercepts have the form (x,0); set y = 0 in the equation and solve for x.

*y-intercepts have the form (0,y); set x = 0 in the equation and solve for y


\begin{align}
y & = 2\sqrt{{x + 4}} 
 -2
\end{align}
Find the X and Y intercepts by setting each variable to 0. I believe I have X right but not Y, please explain where I am messing this up, should be a basic algebra folly.
X intercept: (I believe I have this right, though I should have just subtracted out the -2 early on)
\begin{align}
y & = 2\sqrt{{x + 4}} 
 -2
\end{align}
\begin{align}
0 & = 2\sqrt{{x + 4}} 
 -2
\end{align}
\begin{align}
0^2 & = 2^2{{(x + 4)}} 
 -2^2
\end{align}
\begin{align}
0 & = 4{{x + 16}} 
 -4
\end{align}
\begin{align}
0 & = 4{{x + 12}} 
 \end{align}
\begin{align}
-12 & = 4{{x}} 
 \end{align}
\begin{align}
-3 & = {{x}} 
 \end{align}
Thus the X intercept is (-3, 0)
Y intercept: I know this is wrong, textbook lists the answer as 2
\begin{align}
y & = 2\sqrt{{x + 4}} 
 -2
\end{align}
\begin{align}
y & = 2\sqrt{{0 + 4}} 
 -2
\end{align}
\begin{align}
y & = 2\sqrt{{4}} 
 -2
\end{align}
(Square both sides to remove the radix?)
\begin{align}
y^2 & = 4*4
 -4
\end{align}
\begin{align}
y^2 & = 12
\end{align}
\begin{align}
y & = \sqrt12
\end{align}
Actual answer is 0, 2 for y intercept. So what's my mistake?

For the $x$-intercept: You need to isolate the square root before you square both sides.  So when you have
$$ 0 = 2\sqrt{x+4}-2,$$
you don't want to square both sides here because you will have to FOIL out the right side like
$$0^{2} = (2\sqrt{x+4}-2)^{2}$$
and you will end up with a mess.
If you get the square root by itself you will have
$$1 = \sqrt{x+4}$$
then you can square both sides to get
$$1^{2} = (\sqrt{x+4})^{2}$$
and you can solve from here. (Notice that you had the same answer in your attempt, but that was pure coincidence)
For the $y$-intercept: You don't need to square both sides, you have
$$y = 2\sqrt{4} -2$$
and you know that $\sqrt{4}=2$, so this is just
$$y=2\cdot2-2=4-2=2.$$
