Graph the function, and compare to the parent graph. State the domain and range.

$\mathit{g}\left(x\right)\mathbf{=}\sqrt{\mathbf{x}}\mathbf{-}\mathbf{4}$

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: $g\left(x\right)=\sqrt{x}-4$

In order to graph a function, find few co-ordinates by substituting values of ‘$x$’ and find finding the respective values of ‘$g\left(x\right)$'.

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0,\\ g\left(0\right)=\sqrt{0}-4\\ =0-4\\ =-4\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=1,\\ g\left(1\right)=\sqrt{1}-4\\ =1-4\\ =-3\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=4,\\ g\left(4\right)=\sqrt{4}-2\\ =2-4\\ =-2\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ g\left(9\right)=\sqrt{9}-4\\ =3-4\\ =-1\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=16,\\ g\left(x\right)=\sqrt{16}-4\\ =4-4\\ =0\end{array}$

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

The parent function of$g\left(x\right)=\sqrt{x}-4$ is the simple square root function.

That is,$g\left(x\right)=\sqrt{x}$

The graph of parent function$g\left(x\right)=\sqrt{x}$ is given below.

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

The graph$g\left(x\right)=\sqrt{x}-4$ is obtained by parent function is subtracted by ‘4’.

Therefore, the graph$g\left(x\right)=\sqrt{x}-4$ is translated(shifted) downward by 4 units from the origin, on comparing with the parent graph $g\left(x\right)=\sqrt{x}$.

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

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

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

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

In$g\left(x\right)=\sqrt{x}-4$ the square root of xis subtracted by ‘4’.

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

Find the starting value of the function by substituting $x=0$in $g\left(x\right)=\sqrt{x}-4$.

$\begin{array}{c}g\left(0\right)=\sqrt{0}-4\\ =0-4\\ =-4\end{array}$

Also, in $g\left(x\right)=\sqrt{x}-4$, coefficient of $\sqrt{x}$is 1, which is positive.

Therefore, $g\left(x\right)$takes all the real values greater than or equal to ‘$-4$’.

That is,$g\left(x\right)\ge -4,\Rightarrow g\left(x\right)\in [-4,\infty )$

Therefore, Range:$[-4,\infty )$