## Precalculus (6th Edition) Blitzer

Consider the equation $y=4-{{x}^{2}}$. Substitute $x=-3$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{\left( -3 \right)}^{2}} \\ & =4-9 \\ & =-5 \end{align} Substitute $x=-2$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{\left( -2 \right)}^{2}} \\ & =4-4 \\ & =0 \end{align} Substitute $x=-1$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{\left( -1 \right)}^{2}} \\ & =4-1 \\ & =3 \end{align} Substitute $x=0$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{0}^{2}} \\ & =4-0 \\ & =4 \end{align} Substitute $x=1$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{1}^{2}} \\ & =4-1 \\ & =3 \end{align} Substitute $x=2$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{2}^{2}} \\ & =4-4 \\ & =0 \end{align} Substitute $x=3$ and solve: \begin{align} & y=4-{{x}^{2}} \\ & =4-{{3}^{2}} \\ & =4-9 \\ & =-5 \end{align}