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

$\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}\mathbf{+}\mathbf{3}$

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: $y=\sqrt{x}+3$

In order to graph a function, find few co-ordinates by substituting values of ‘$x$’ and find finding the respective values of ‘localid="1648039170401" $y$’.

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0,\\ y=\sqrt{0}+3\\ =0+3\\ =3\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=1,\\ y=\sqrt{1}+3\\ =1+3\\ =4\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=4,\\ y=\sqrt{4}+3\\ =2+3\\ =5\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ y=\sqrt{9}+3\\ =3+3\\ =6\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=16,\\ y=\sqrt{16}+3\\ =4+3\\ =7\end{array}$

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

The parent function of$y=\sqrt{x}+3$ is the simplest square root function.

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

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 graph$y=\sqrt{x}+3$ is obtained by adding ‘3’ to the parent function.

Therefore, the graph$y=\sqrt{x}+3$ is translated(shifted) upward by 3 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 $y=\sqrt{x}+3$ the square root of $x$is added by ‘3’.

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

Find the starting value of $y$by substituting $x=0$in $y=\sqrt{x}+3$.

$\begin{array}{c}y=\sqrt{0}+3\\ =0+3\\ =3\end{array}$

Also, in $y=\sqrt{x}+3$, coefficient of $\sqrt{x}$is 1, which is positive.

Therefore,$y$ takes all the real values greater than or equal to ‘3’.

That is,$y\ge 3,\Rightarrow y\in [3,\infty )$

Therefore, Range:$[3,\infty )$