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.