## Precalculus (6th Edition) Blitzer

Let us consider the inequalities; put the equals symbol in place of the inequality and rewrite the equation as given below: $y=1-{{x}^{2}}$ ${{x}^{2}}+{{y}^{2}}=9$. To plot the graph of $y=1-{{x}^{2}}$, evaluate the value of the y variable for each value of the x variable. To find the value of the y-intercept, substitute $x = 0$ as given below: \begin{align} & y=1-{{\left( 0 \right)}^{2}} \\ & =1-0 \\ & =1 \end{align} To find the value of the x-intercept, substitute $y = 0$ as given below: \begin{align} & 0=1-{{x}^{2}} \\ & {{x}^{2}}=1 \\ & x=\pm 1 \end{align} Thus, plot the intercepts $\left( 0,1 \right),\ \left( 1,0 \right)\text{, and }\left( -1,0 \right)$ and join them with a free hand in order to obtain the graph of the equation $y=1-{{x}^{2}}$. Then, this parabola divides the plane into three regions: the parabola itself, the inner plane of the parabola, and the outer plane of the parabola. Now, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade: \begin{align} & y\le 1-{{x}^{2}} \\ & 0\overset{?}{\mathop{\le }}\,1-0 \\ & 0\le 1 \\ \end{align} Since the test point satisfies the inequality, shade the inner plane of the parabola, that is, towards the origin $\left( 0,0 \right)$. For a circle having center $\left( h,k \right)$ and radius $r$, the general equation of circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$. And the second equation ${{x}^{2}}+{{y}^{2}}=9$ represents a circle with origin at $\left( 0,0 \right)$ and radius 3 that can be rewritten as ${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{3}^{2}}$. Plot the graph using the equation of this circle in the rectangular coordinate system with origin at $\left( 0,0 \right)$ and radius $r=3$. Now, this circle divides the plane into three regions: the circle itself, the inner plane of the circle, and the outer plane of the circle. Then, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade: \begin{align} & {{x}^{2}}+{{y}^{2}}\le 9 \\ & 0+0\overset{?}{\mathop{\le }}\,9 \\ & 0\le 9 \\ \end{align} . Since the test point satisfies the inequality, shade the inner plane of the circle, that is, towards the origin $\left( 0,0 \right)$. See the final graph below.