## Precalculus (6th Edition) Blitzer

We have to plot the graph of the functions with the help of the points individually for $f\left( x \right)$ and $g\left( x \right)$. The co-ordinates of $f\left( x \right)$ for $x=0,1,4,9$ are as follows: Put $x=0$ \begin{align} & f\left( x \right)=\sqrt{x} \\ & =0 \end{align} Put $x=1$ \begin{align} & f\left( x \right)=\sqrt{x} \\ & =\sqrt{1} \\ & =1 \end{align} Put $x=4$ \begin{align} & f\left( x \right)=\sqrt{x} \\ & =\sqrt{4} \\ & =2 \end{align} Put $x=9$ \begin{align} & f\left( x \right)=\sqrt{x} \\ & =\sqrt{9} \\ & =3 \end{align} The co-ordinates of $g\left( x \right)=\sqrt{x+2}$ for $x=-2,-1,2,7$ are as follows: Put $x=-2$ \begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{-2+2} \\ & =0 \end{align} Put $x=-1$ \begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{-1+2} \\ & =\sqrt{1} \\ & =1 \end{align} Put $x=2$ \begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{2+2} \\ & =\sqrt{4} \\ & =2 \end{align} Put $x=7$ \begin{align} & g\left( x \right)=\sqrt{x+2} \\ & =\sqrt{7+2} \\ & =\sqrt{9} \\ & =3 \end{align}