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

Aug 8, 2017

Crosses at $y = \pm 4$

#### Explanation:

As this equation is already in the correct format, put $x \mathmr{and} y$ 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: $\frac{x}{a} = \pm \frac{y}{b}$ to determine the equations of the asymptotes where $a = \sqrt{9} , b = \sqrt{16}$ and you will get $y = \pm \left(\frac{4}{3}\right) x$