Graph the function. Compare to the parent graph. State the domain and range.

$\mathit{y}\mathbf{=}\sqrt{\mathbf{x}\mathbf{+}\mathbf{2}}$

Parent graph:The simplest form of the given function is called the parent function of that function and the graph of the parent function is called parent graph.

Domain: The set of all possible values for which given function defined is called domain.

Range: The set of all possible values of the given function is called range.

The given function is: $y=\sqrt{x+2}$

In order to graph a function, find few co-ordinates by substituting values of ‘$x$’ and find finding the respective values of ‘$y$’.

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=-2,\\ y=\sqrt{-2+2}\\ =\sqrt{0}\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=-1,\\ y=\sqrt{-1+2}\\ =\sqrt{1}\\ =1\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=2,\\ y=\sqrt{2+2}\\ =\sqrt{4}\\ =2\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=7,\\ y=\sqrt{7+2}\\ =\sqrt{9}\\ =3\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=14,\\ y=\sqrt{14+2}\\ =\sqrt{16}\\ =4\end{array}$

Plot these co-ordinates on a coordinate plane and join those points to get the required graph.

The parent function of $y=\sqrt{x+2}$is the simplest square root function.

That is, $y=\sqrt{x}$

The graph of parent function $y=\sqrt{x}$is given below.

Note: Since the parent function is just used for comparison, it is graphed using graphing calculator.

The value 2 is added inside square root of parent function $y=\sqrt{x}$. So the graph$y=\sqrt{x+2}$ is the translation of the parent graph $y=\sqrt{x}$.

Since 2 is added inside the root, the graph is translated to the left by 2 units.

Since the values inside the root must be positive.

$x+2\ge 0$

Adding ‘$-2$’ on both the sides.

$\begin{array}{c}x+2-2\ge 0-2\\ x+0\ge -2\\ x\ge -2\end{array}$

$x\ge -2\Rightarrow x\in [-2,\infty )$

Therefore, domain: $[-2,\infty )$

The $y$values are square root of $x+2$. As square roots are positive$y$ takes all the positive real values including zero.

Therefore,$y\ge 0,\Rightarrow y\in [0,\infty )$

Therefore, Range:$[0,\infty )$