Question

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

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

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 the 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 the parent graph.

Step 2. State the concept of domain and range.

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.

Step 3. Graph the function.

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 ‘$y$’.

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

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

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

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

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

Values of ‘$x$	Values of ‘$y$'	$\left(x,y\right)$
0	2	(0,2)
1	3	(1,3)
4	4	(4,4)
9	5	(9,5)
16	6	(16,6)

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

Step 4. Comparison with the parent 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 $y=\sqrt{x}$.

Step 5. State the domain and range.

Since ‘$x$’ is inside the root, the values inside the root must be positive.

Therefore, values of$x$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:$[2,\infty )$