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?

In the first part, $\begin{align}
0^2 & \ne 2^2{{(x + 4)}} 
 -2^2
\end{align}$ In general, $(a-b)^2\ne a^2-b^2$
Similar problem in the second part. Just evaluate square root 4 (hint it's 2) and avoid the following missteps.
