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

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

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

To graph a function, find a few coordinates by substituting values of ‘$x$’ and find finding the respective values of ‘$y$’.

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0,\\ y=\sqrt{0}+2\\ =0+2\\ =2\end{array}$

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

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

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ y=\sqrt{9}+2\\ =3+2\\ =5\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=16,\\ y=\sqrt{16}+2\\ =4+2\\ =6\end{array}$

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

The parent function $y=\sqrt{x}+2$is the simplest 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$y=\sqrt{x}+2$ is obtained by adding ‘2’ to the parent function.

Therefore, the graph$y=\sqrt{x}+2$ 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$x$are all positive real numbers including zero.

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

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

In $y=\sqrt{x}+2$ the square root of $x$is added by ‘2’.

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 $y=\sqrt{x}+2$.

$\begin{array}{c}y=\sqrt{0}+2\\ =0+2\\ =2\end{array}$

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

Therefore,$y$ takes all the real values greater than or equal to ‘2’.

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

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