Question

How do you graph the hyperbola `y^2/16-x^2/9=1`?

Answer 1

Crosses at `y=+-4`

Explanation:

As this equation is already in the correct format, put `x and y` equal to zero and you will see that when `x=0`, `y^2=16` which gives the y intercepts.

If you put `y=0`, you get `x^2=-9` which doesn't give any real roots. The graph will therefore be a hyperbola in the shape of a V and an upside down V (as opposed to being on their sides like ><).

You can use the formula: `x/a = +-y/b` to determine the equations of the asymptotes where `a=sqrt(9), b=sqrt(16)` and you will get `y=+-(4/3)x`