Answer
x-intercepts: $(-4,0),(4,0)$
y-intercepts: $(0,-2),(0,2)$
Work Step by Step
The intercepts of the equation $x^2 + 4y^2 = 16$ are:
For $x$-intercepts, set $y=0$:
$x^{2}+4(0)^{2}=16$
$x^{2}=16$
$x=\pm 4$
$(-4,0),(4,0)$
For $y$-intercepts, set $x=0$:
$(0)+4y^{2}=16$
$4y^{2}=16$
$y^{2}=4$
$y=\pm 2$
$(0,-2),(0,2)$
