## Calculus 10th Edition

The intercepts with $y$ axis are $y=1$ and $y=-1$. The intercepts with $x$ axis are $x=2$ and $x=-2$. The graph is symmetric with respect to both $x$ and $y$ axis and is shown on the figure.
To find the intercepts with $y$ axis put $x=0$ and calculate $y$: $$0^2+4y^2 = 4\Rightarrow 4y^2=4\Rightarrow y^2=1\Rightarrow y=\pm 1$$ so there are two intercepts with $y$ axis $y=1$ and $y=-1$. To find the intercept with $x$ axis put $y=0$ and calculate $x$: $$x^2+4\times0^2=4\Rightarrow x^2=4\Rightarrow x=\pm2$$ so there are also two intercepts with the $x$ axis $x=2$ and $x=-2$. The graph is symmetric with respect to both $y$ and $x$ axis because if we put $-y$ instead of $y$ or $-x$ instead of $x$ we obtain the starting equation: $$(-x)^2+4y^2=x^2+4y^2=4.$$ $$x^2+4(-y)^2 = x^2+4y^2=4.$$