Question

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

Answer 1

See the explanation below

Explanation:

The domain of `y` is `D_y=RR-{0}`

As `x!=0`

So, `x=0` is a vertical asymptote.

The numerator is `x^2-5x-36`

`=(x+4)(x-9)`

So, the intercepts when `y=0`

are `(-4,0)` and `(9,0)`

For the limits `x->+-oo`, we take the terms of highest degree in the numerator and the denominator

`lim_(x->+-oo)y=lim_(x->+-oo)x^2/(3x)=lim_(x->+-oo)x/3=+-oo`

`lim_(x->0^(-))y=(0^(+)+5-36)/(3*0^(-))=+oo`

`lim_(x->0^(+))y=(0^(+)-5-36)/(3*0^(+))=-oo`

Now you can draw your graph

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

Answer 2

Asymptotes: `x = 0 and x-3y-5/3=0.` x-intercepts by the two branches: -4 and 9. As `x to 0, y to +-oo` and the other asymptote tends to reach the hyperbola, at infinite distance in `Q_1 and Q_3`

Explanation:

graph{x^2-3xy-5x-36=0 [-20, 20, -10, 10]} The second degree equation `ax^2+2hxy+by^2+...=0`

represents a hyperbola, when `ab-h^2<0` and the second degree

terms reveal the first degree terms in the equations of its

asymptotes.

Here, after cross multiplication,

`x(x-3y)-5x-36=0`. Reconstructing as

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

`a =0, b=-5/3 and c=-36`.

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 `x to 0, y to +-oo` and the other asymptote tends to reach the

hyperbola, at infinite distance in `Q_1 and Q_3`