College Algebra, Chapter 2, 2.2, Section 2.2, Problem 26

Make a table of values and sketch the graph of the equation $x^2 + y^2 = 9$. Find the $x$ and $y$ intercepts.

$
\begin{array}{|c|c|}

\hline\\
\text{Let } x & y = \sqrt{9 - x^2} \\
\hline\\
-3 & 0 \\
\hline\\
-2 & \sqrt{5} \\
\hline\\
-1 & \sqrt{8} \\
\hline\\
1 & \sqrt{8} \\
\hline\\
2 & \sqrt{5} \\
\hline\\
3 & 0\\
\hline

\end{array} $

To solve for $x$ intercept, where $y = 0$


$
\begin{equation}
\begin{aligned}

x^2 + 0^2 =& 9
\\
\\
x^2 =& 9
\\
\\
x =& \pm \sqrt{9} = \pm 3

\end{aligned}
\end{equation}
$


Thus, the $x$ intercept is at $(3,0)$ and $(-3, 0)$

To solve for the $y$ intercept, we set $x = 0$


$
\begin{equation}
\begin{aligned}

0 + y^2 =& 9
\\
\\
y =& \sqrt{9} = 3

\end{aligned}
\end{equation}
$


Thus, the $y$ intercept is at $(0, 3)$