# How do you graph #y=(x^2-5x-36)/(3x)# using asymptotes, intercepts, end behavior?

##### 2 Answers

See the explanation below

#### Explanation:

The domain of

As

So,

The numerator is

So, the intercepts when

are

For the limits

Now you can draw your graph

graph{(x^2-5x-36)/(3x) [-32.47, 32.47, -16.24, 16.25]}

Asymptotes:

#### Explanation:

graph{x^2-3xy-5x-36=0 [-20, 20, -10, 10]} The second degree equation

represents a hyperbola, when

terms reveal the first degree terms in the equations of its

asymptotes.

Here, after cross multiplication,

((x+a)(x-3y+b)+c=0, we find that

So, the asymptotes are given by

x=0 and x-3y-5/3=0, meeting at the center C(0, -5/9)#.

x-intercepts by the two branches: -4 and 9.

As

hyperbola, at infinite distance in