Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2 - 9)/(3x-6)`?

Answer 1

**Vertical Asymptote: ** `color(blue)(x = 2)`

**Horizontal Asymptote: ** None

Equation of the Slant/Oblique Asymptote: `color(blue)(y = x/3+2/3)`

Explanation:

Given:

`color(red)(y = f(x) = (x^2-9)/(3x-6))`

`color(green)(Step.1):`

To find the Vertical Asymptote:

a. Factor where possible

b. Cancel common factors, if any

c. Set Denominator = 0

We will start following the steps:

Consider:

`color(red)(y = f(x) = (x^2-9)/(3x-6))`

We will factor where possible:

`y = f(x) = [(x+3)(x-3)]/(3x-6)`

If there are any common factors in the numerator and the denominator, we can cancel them.

But, we do not have any.

Hence, we will move on.

Next, we set the denominator to zero.

`color(blue)((3x-6) = 0`

Add `6` to both sides.

`(3x-6+6) = 0+6`

`(3x-cancel 6+cancel 6) = 0+6`

`rArr 3x = 6`

`rArr x = 6/3 = 2`

Hence, our Vertical Asymptote is at `color(blue)(x=2)`

Refer to the graph below:

`color(green)(Step.2):`

To find the Horizontal Asymptote:

Consider:

`color(red)(y = f(x) = (x^2-9)/(3x-6))`

Since the highest degree of the numerator is greater than the highest degree of the denominator,

Horizontal Asymptote DOES NOT EXIST

`color(green)(Step.3):`

To find the Slant/Oblique Asymptote:

Consider:

`color(red)(y = f(x) = (x^2-9)/(3x-6))`

Since, the highest degree of the numerator is one more than the highest degree of the denominator, we do have a Slant/Oblique Asymptote

We will now perform the Polynomial Long Division using
`color(red)(y = f(x) = (x^2-9)/(3x-6))`

Hence, the Result of our Long Polynomial Division is

`x/3 + 2/3 + (-5/(3x-6))`

Equation of the Slant/Oblique Asymptote is

`color(blue)(y = x/3 + 2/3)`

Refer to the graph below:

We have all the required results now.