Question

How do you find all the critical points to graph `(x/2)^2 - (y/3)^2 = 1` including vertices, foci and asymptotes?

Answer 1

All the answers are below

Explanation:

Th graph{x^2/4-y^2/9=1 [-10, 10, -5, 5]} is is a hyperbola
The center is `(0,0)` at the origin
The vertices are `(2,0)` and `(-2,0)`
The slopes of the asymptotes are `(3/2)` and `(-3/2)`
The equations of the asymptotes are `y=(3x)/2` and `y=(-3x)/2`
To determine the foci, we calculate, `c=+-sqrt(2^2+3^2)=+-sqrt13`
So the foci are `(sqrt13,0)` and `(-sqrt13,0)`