## College Algebra (10th Edition)

$$\\ (2,0) \ \text{and} \ (-2, 0)$$ $$\\(0, 4) \ \text{and} \ (0, -4)$$
$$y^2=16-4x^2$$ Set y=0 to find the x intercept: $$16-4x^2=0 \\x^2=4 \\x= \pm 2 \\ (2,0) \\ \text{and} \\ (-2, 0)$$ Set x=0 to find the y intercept: $$y^2=16 \\y= \pm 4 \\(0, 4) \ \text{and} \ (0, -4)$$