## Precalculus (10th Edition)

$(-4,0),(4,0),(0,-2), \text{ and }(0,2)$.
The given equation is $x^2+4y^2=16$ To find the $x$ intercepts, set $y=0$, then solve for $x$. $x^2+4(0)^2=16$ $x^2=16$ $x=\pm 4$ Hence the $x$ intercepts are $(-4,0),(4,0)$. To find the $y$ intercepts, set $x=0$, then solve for $y$ $(0)^2+4y^2=16$ $4y^2=16$ $y^2=4$ $y=\pm 2$ Hence the $y$ intercepts are $(0,-2),(0,2)$