How would I make a graph for $\sqrt{x+1}-3$ I have to make a table for $\sqrt{x+1}-3$ and I can't figure out how to find the $x$ values.  I know that I have to get the middle section to equal perfect squares, which are $0,\ 1,\ 4,\ 9$ but I don't know how. Please help!
Is it $2,\ 3,\ 6$ and $12$? 
The problem is when I get the $x$ values, the $y$ values are in decimal form. Are they supposed to be?
$$\begin{array}{|c|c|c|}
\hline
x & \sqrt{x+1}-3 & y\\
\hline
? & & \\
\hline
? & & \\
\hline? & & \\
\hline? & & \\
\hline? & & \\
\hline
\end{array}$$

To obtain integer values for $y = \sqrt{x + 1} - 3$, $x + 1$ must be a perfect square.  The first five perfect squares are $0, 1, 4, 9, 16$.  Thus, we must set $x + 1$ equal to $0, 1, 4, 9, 16$.
If $x + 1 = 0$, then $x = -1$.  If $x = -1$, then $y = \sqrt{-1 + 1} - 3 = \sqrt{0} - 3 = -3$.  Hence, the point $(-1, -3)$ is on the graph.  
If $x < -1$, then $x + 1 < 0$, so $\sqrt{x + 1}$ is not a real number.  Hence, $(-1, 0)$ is the endpoint of the graph.
If $x + 1 = 1$, then $x = 0$.  If $x = 0$, then $y = \sqrt{0 + 1} - 3 = \sqrt{1} - 3 = 1 - 3 = -2$.  Hence, the point $(0, -2)$ is on the graph.
If $x + 1 = 4$, then $x = 3$.  If $x = 3$, then $y = \sqrt{3 + 1} - 3 = \sqrt{4} - 3 = 2 - 3 = -1$.  Hence, the point $(3, -1)$ is on the graph.
As you can check, setting $x + 1 = 9$ yields the point $(8, 0)$ and setting $x + 1 = 16$ yields the point $(15, 1)$.
You can plot these points on the coordinate plane and connect them with a smooth curve to obtain the graph.

Note that the graph of $y = \sqrt{x + 1} - 3$, which is shown in green, can be obtained from the graph of $y = \sqrt{x}$, which is shown in blue, by shifting the graph of $y = \sqrt{x}$ to the left by one unit and down by three units.  
