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

$\mathit{y}\mathbf{=}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}$

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 the parent graph.

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

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

The given function is:$y=\sqrt{x+1}$

To graph a function, find a few coordinates by substituting values of ‘$x$’ and find finding the respective values of ‘$y$’.

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=-1,\\ y=\sqrt{-1+1}\\ =\sqrt{0}\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0,\\ y=\sqrt{0+1}\\ =\sqrt{1}\\ =1\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=3,\\ y=\sqrt{3+1}\\ =\sqrt{4}\\ =2\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=8,\\ y=\sqrt{8+1}\\ =\sqrt{9}\\ =3\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=15,\\ y=\sqrt{15+1}\\ =\sqrt{16}\\ =4\end{array}$

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

The parent function $y=\sqrt{x+1}$is the simplest square root function.

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

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

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

The value 1 is added inside the square root of the parent function $y=\sqrt{x}$. So the graph $y=\sqrt{x+1}$is the translation of the parent graph $y=\sqrt{x}$.

Since 1 is added inside the root, the graph is translated to the left by 1 unit.

Since the values inside the root must be positive.

$x+1\ge 0$

Adding ‘$-1$’ on both sides.

$\begin{array}{c}x+1-1\ge 0-1\\ x+0\ge -1\\ x\ge -1\end{array}$

$x\ge -1\Rightarrow x\in [-1,\infty )$

Therefore, domain: $[-1,\infty )$

The values of $y$ are the square root of $x+1$. 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:$[0,\infty )$