## Precalculus (6th Edition) Blitzer

We plot the graphs of functions with the help of the points individually for $f\left( x \right)$ and $g\left( x \right)$. The coordinates of $f\left( x \right)$ for $x=-2$ to $x=2$ are as follows: \begin{align} & x=-2 \\ & f\left( x \right)=-2x \\ & =-2\left( -2 \right) \\ & =4 \end{align} \begin{align} & x=-1 \\ & f\left( x \right)=-2x \\ & =-2\left( -1 \right) \\ & =2 \end{align} \begin{align} & x=0 \\ & f\left( x \right)=-2x \\ & =-2\left( 0 \right) \\ & =0 \end{align} \begin{align} & x=1 \\ & f\left( x \right)=-2x \\ & =-2\left( 1 \right) \\ & =-2 \end{align} \begin{align} & x=2 \\ & f\left( x \right)=-2x \\ & =-2\left( 2 \right) \\ & =-4 \end{align} The coordinates of $g\left( x \right)=-2x+3$ for $x=-2$ to $x=2$ are as follows: \begin{align} & x=-2 \\ & g\left( x \right)=-2x+3 \\ & =-2\left( -2 \right)+3 \\ & =7 \end{align} \begin{align} & x=-1 \\ & g\left( x \right)=-2x+3 \\ & =-2\left( -1 \right)+3 \\ & =5 \end{align} \begin{align} & x=0 \\ & g\left( x \right)=-2x+3 \\ & =-2\left( 0 \right)+3 \\ & =3 \end{align} \begin{align} & x=1 \\ & g\left( x \right)=-2x+3 \\ & =-2\left( 1 \right)+3 \\ & =1 \end{align} \begin{align} & x=2 \\ & g\left( x \right)=-2x+3 \\ & =-2\left( 2 \right)+3 \\ & =-1 \end{align}