Graph the function. Compare to the parent graph. State the domain and range.

$\mathit{y}\mathbf{=}\mathbf{5}\sqrt{\mathbf{x}}$

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: $y=5\sqrt{x}$

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

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=0,\\ y=5\sqrt{0}\\ =3\left(0\right)\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=1,\\ y=5\sqrt{1}\\ =5\left(1\right)\\ =5\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=4,\\ y=5\sqrt{4}\\ =5\left(2\right)\\ =10\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=9,\\ y=5\sqrt{9}\\ =5\left(3\right)\\ =15\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}x=16,\\ y=5\sqrt{16}\\ =5\left(4\right)\\ =20\end{array}$

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

The parent function of$y=5\sqrt{x}$ is the simple square root function.

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

The graph of parent function$y=5\sqrt{x}$ 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 ‘5’, which is greater than 1.

Therefore, the graph$y=5\sqrt{x}$ is a vertical stretch of 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 $y=5\sqrt{x}$, the square root of$x$ is multiplied by ‘5’.

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 $y=5\sqrt{x}$.

$\begin{array}{c}y=5\left(\sqrt{0}\right)\\ =5\left(0\right)\\ =0\end{array}$

In $y=5\sqrt{x}$, coefficient of $\sqrt{x}$is 5 which is positive.

Therefore, $y$takes all the values in real numbers which are greater than or equal to zero.

Therefore,$y\ge 0,\Rightarrow y\in [0,\infty )$

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