College Algebra, Chapter 2, 2.2, Section 2.2, Problem 28

Make a table of values and sketch the graph of the equation $y = - \sqrt{4 - x^2}$. Find the $x$ and $y$ intercepts.

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

\hline\\
\text{Let } x & y = - \sqrt{4 - x^2} \\
\hline\\
-2 & 0 \\
\hline\\
-1.5 & - \displaystyle \frac{\sqrt{7}}{2} \\
\hline\\
-1 & - \sqrt{3} \\
\hline\\
-0.5 & - \displaystyle \frac{\sqrt{15}}{2} \\
\hline\\
0.5 & - \displaystyle \frac{\sqrt{15}}{2} \\
\hline\\
1 & - \sqrt{3}\\
\hline\\
1.5 & - \displaystyle \frac{\sqrt{7}}{2}\\
\hline\\
2 & 0\\
\hline

\end{array} $

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


$
\begin{equation}
\begin{aligned}

0 =& - \sqrt{4 - x^2}
\\
\\
0 =& \sqrt{4 - x^2}
\\
\\
0 =& 4 - x^2
\\
\\
x^2 =& 4
\\
\\
x =& \pm 2


\end{aligned}
\end{equation}
$


Thus, the $x$ intercept is at $(2,0)$ and $(-2, 0)$

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


$
\begin{equation}
\begin{aligned}

y =& - \sqrt{4 - (0)^2}
\\
\\
y =& - \sqrt{4}
\\
\\
y =& -2

\end{aligned}
\end{equation}
$


Thus, the $y$ intercept is at $(0, -2)$