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 ‘$y$’.

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

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

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

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ y=\sqrt{9}-1\\ =3-1\\ =2\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=16,\\ y=\sqrt{16}-1\\ =4-1\\ =3\end{array}$

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 simple square root function.

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

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 graph$y=\sqrt{x}-1$ is obtained by the parent function is subtracted by ‘1’.

Therefore, the graph$y=\sqrt{x}-1$ is translated(shifted) downward by 1 unit from the origin, on comparing with the parent graph $y=\sqrt{x}$.

Since ‘$x$’ is inside the root, the values inside the root must be positive.

Therefore, values of$x$are all positive real numbers including zero.

That is, $x\ge 0,\Rightarrow x\in [0,\infty )$.

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

The square root of $x$is subtracted by ‘1’.

As the square root is always positive, the least value it takes is zero.

Find the starting value of $y$ by substituting $x=0$in $y=\sqrt{x}-1$.

$\begin{array}{c}y=\sqrt{0}-1\\ =0-1\\ =-1\end{array}$

Also, in $y=\sqrt{x}-1$, coefficient of $\sqrt{x}$is 1, which is positive.

Therefore,y takes all the real values greater than or equal to ‘$-1$’.

That is,$y\ge -1,\Rightarrow y\in [-1,\infty )$

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