College Algebra, Chapter 2, 2.2, Section 2.2, Problem 24

Make a table of values and sketch the graph of the equation $x + y^2 = 4$. Find the $x$ and $y$ intercepts.

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


\hline\\
\text{Let } x & y = \sqrt{4 - x} & y = -\sqrt{4-x} \\
\hline\\
-4 & \sqrt{8} & -\sqrt{8}\\
\hline\\
-3 & \sqrt{7} & -\sqrt{7}\\
\hline\\
-2 & \sqrt{6} & -\sqrt{6}\\
\hline\\
-1 & \sqrt{5} & -\sqrt{5}\\
\hline\\
1 & \sqrt{3} & -\sqrt{3} \\
\hline\\
2 & \sqrt{2} & -\sqrt{2} \\
\hline\\
3 & 1 & -1\\
\hline\\
4 & 0 & 0\\
\hline

\end{array} $

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


$
\begin{equation}
\begin{aligned}

x + (0)^2 =& 4
\\
\\
x =& 4

\end{aligned}
\end{equation}
$


Thus, the $x$ intercept is at $(4,0)$

To solve for the $y$ intercepts on both sides of equation, we set $x = 0$


$
\begin{equation}
\begin{aligned}

0 + y^2 =& 4
\\
\\
y^2 =& 4
\\
\\
y =& pm \sqrt{4} = \pm 2

\end{aligned}
\end{equation}
$


Thus, the $y$ intercepts are at $(0, 2)$ and $(0,-2)$