How do you graph #y^2/16-x^2/4=1# and identify the foci and asympototes?
1 Answer
Oct 22, 2016
The foci are
And the asymptotes are
Explanation:
The graph is a hyperbola "opens up and down"
The center is
The vertices are
The slopes of the asymptotes are
The equations of the asymptotes are
To determine the foci, we need
The foci are
graph{(y/4)^2-(x/2)^2=1 [-40, 40, -20, 20]}