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

$\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{\mathbf{x}}\mathbf{-}\mathbf{1}$

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 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: $y=\frac{1}{2}\sqrt{x}-1$

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

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0,\\ y=\frac{1}{2}\sqrt{0}-1\\ =\frac{1}{2}\left(0\right)-1\\ =-1\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=1,\\ y=\frac{1}{2}\sqrt{1}-1\\ =\frac{1}{2}\left(1\right)-1\\ =0.5-1\\ =-0.5\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=4,\\ y=\frac{1}{2}\sqrt{4}-1\\ =\frac{1}{2}\left(2\right)-1\\ =1-1\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ y=\frac{1}{2}\sqrt{9}-1\\ =\frac{1}{2}\left(3\right)-1\\ =1.5-1\\ =0.5\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=16,\\ y=\frac{1}{2}\sqrt{16}-1\\ =\frac{1}{2}\left(4\right)-1\\ =2-1\\ =1\end{array}$

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

The parent function of$y=\frac{1}{2}\sqrt{x}-1$ is the simplest square root function.

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

Using graphing calculator, 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 parent function is multiplied by$\frac{1}{2}$ (a value less than 1 and greater than zero) and is subtracted is ‘1’. So the graph$y=\frac{1}{2}\sqrt{x}-1$ is a vertical compression of$y=\sqrt{x}$ and is translated(shifted) downward by 1 units 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$x$is all positive real numbers including zero.

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

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

In $y=\frac{1}{2}\sqrt{x}-1$, the square root of$x$is multiplied by$\frac{1}{2}$ and is subtracted by ‘1’.

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 $y=\frac{1}{2}\sqrt{x}-1$

$\begin{array}{c}y=\frac{1}{2}\left(0\right)-1\\ =0-1\\ =-1\end{array}$

In $y=\frac{1}{2}\sqrt{x}-1$, coefficient of $\sqrt{x}$is 1, which is positive.

Therefore $y$takes all the values of real numbers which are greater

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

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