## College Algebra (10th Edition)

x-intercepts: $(-4,0),(4,0)$ y-intercepts: $(0,-2),(0,2)$
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 $x=0$: $(0)+4y^{2}=16$ $4y^{2}=16$ $y^{2}=4$ $y=\pm 2$ $(0,-2),(0,2)$