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.