# How do you graph #x^2/64-y^2=1# and identify the foci and asympototes?

##### 1 Answer

May 27, 2017

For a hyperbola of the form:

The foci are at

The asymptotes are:

#### Explanation:

Matching the given equation,

with equation [1], we observe that

With these values the foci are at:

Simplify:

The asymptotes are:

Simplify:

The graph is:

graph{(x-0)^2/8^2-(y-0)^2/1^2=1 [-10, 10, -5, 5]}