Precalculus, Chapter 1, 1.2, Section 1.2, Problem 66

Find the intercepts of the equation $y = x^2 + 4$ and test for symmetry.

$x$-intercepts:


$
\begin{equation}
\begin{aligned}

y =& x^2 + 4
&& \text{Given equation}
\\
0 =& x^2 + 4
&& \text{To find the $x$-intercept, we let } y = 0
\\
-4 =& x^2
&&
\\
\sqrt{-4} =& x
&&

\end{aligned}
\end{equation}
$


There is no real solutions for $x$

$y$-intercepts:


$
\begin{equation}
\begin{aligned}

y =& x^2 + 4
&& \text{Given equation}
\\
y =& (0)^2 + 4
&& \text{To find the $y$-intercept, we let } x = 0
\\
y =& 4
&&

\end{aligned}
\end{equation}
$


The $y$-intercept is $(0,4)$

Test for symmetry

$x$-axis:


$
\begin{equation}
\begin{aligned}

y =& x^2 + 4
&& \text{Given equation}
\\
-y =& x^2 + 4
&& \text{To test for $x$-axis symmetry, replace $y$ by $-y$ and see if the equation is still the same}

\end{aligned}
\end{equation}
$


The equation changes so it is not symmetric to the $x$-axis

$y$-axis:


$
\begin{equation}
\begin{aligned}

y =& x^2 + 4
&& \text{Given equation}
\\
y =& (-x)^2 + 4
&& \text{To test for $y$-axis symmetry, replace$ x$ by $-x$ and see if the equation is still the same}
\\
y =& x^2 + 4
&&

\end{aligned}
\end{equation}
$


The equation is still the same so it is symmetric to the $y$-axis

Origin:


$
\begin{equation}
\begin{aligned}

y =& x^2 + 4
&& \text{Given equation}
\\
-y =& (-x)^2 + 4
&& \text{To test for origin symmetry, replace both $x$ by $-x$ and y by $-y$ and see if the equation is still the same}
\\
-y =& x^2 + 4
&&

\end{aligned}
\end{equation}
$


The equation changes so it is not symmetric to the origin.

Therefore, the equation $y = x^2 + 4$ has an intercepts $(0,4)$ and it is symmetric to the $y$-axis.