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

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

$\begin{array}{c}\text{¹ó´Ç°ù â¶Ä‰}x=0,\\ y=\frac{1}{2}\sqrt{0}+2\\ =\frac{1}{2}\left(0\right)+2\\ =2\end{array}$

$\begin{array}{c}\text{¹ó´Ç°ù â¶Ä‰}x=1,\\ y=\frac{1}{2}\sqrt{1}+2\\ =\frac{1}{2}\left(1\right)+2\\ =0.5+2\\ =2.5\end{array}$

$\begin{array}{c}\text{¹ó´Ç°ù â¶Ä‰}x=4,\\ y=\frac{1}{2}\sqrt{4}+2\\ =\frac{1}{2}\left(2\right)+2\\ =1+2\\ =3\end{array}$

$\begin{array}{c}\text{¹ó´Ç°ù â¶Ä‰}x=9,\\ y=\frac{1}{2}\sqrt{9}+2\\ =\frac{1}{2}\left(3\right)+2\\ =1.5+2\\ =3.5\end{array}$

$\begin{array}{c}\text{¹ó´Ç°ù â¶Ä‰}x=16,\\ y=\frac{1}{2}\sqrt{16}+2\\ =\frac{1}{2}\left(4\right)+2\\ =2+2\\ =4\end{array}$

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

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

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, $xâ‰\yen 0,\text{ â¶Ä‰â¶Ä‰}⇒x∈[0,∞)$.

Therefore, domain: $[0,∞)$

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, $yâ‰\yen 2,\text{ â¶Ä‰â¶Ä‰}⇒y∈[2,∞)$

Therefore, Range: $[2,∞)$