Question

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

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

Short answer

The domain of the given function is$\mathit{x}\mathbf{\in }\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ and the range is $\mathit{y}\mathbf{\in }\mathbf{[}\mathbf{2}\mathbf{,}\mathbf{\infty }\mathbf{)}$.

Step-by-step solution

Step 1. State the concept of parent graph.

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.

Step 2. State the concept of domain and range.

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.

Step 3. Graph the function.

The given function is: $y=\frac{1}{2}\sqrt{x}+2$

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

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

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=1,\\ y=\frac{1}{2}\sqrt{1}+2\\ =\frac{1}{2}\left(1\right)+2\\ =0.5+2\\ =2.5\end{array}$

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

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ y=\frac{1}{2}\sqrt{9}+2\\ =\frac{1}{2}\left(3\right)+2\\ =1.5+2\\ =3.5\end{array}$

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

Values of ‘$x$	Values of ‘$y$’	$\left(x,y\right)$
0	2	(0,2)
1	2.5	(1,2.5)
4	3	(4,3)
9	3.5	(9,3.5)
16	4	(16,4)

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

Step 4. Comparison with the parent graph.

The parent function of$y=\frac{1}{2}\sqrt{x}+2$ 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 added by‘2’.

So the graph$y=\frac{1}{2}\sqrt{x}+2$ is a vertical compression of$y=\sqrt{x}$ and is translated (shifted) upward by 2 units from the origin, on comparing with the parent graph $y=\sqrt{x}$.

Step 5. State the domain and range.

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

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

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

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

Therefore$y$ takes all the values of real numbers which are greater than or equal to ‘2’.

That is, $y\ge 2,\Rightarrow y\in [2,\infty )$

Therefore, Range: $[2,\infty )$