Question

How do you find vertical, horizontal and oblique asymptotes for `f(x)= (5x^2-17x-12)/(2x^2-7x-4)`?

Answer 1

The horizontal asymptote is `y=5/2`, the vertical asymptote is `x=-1/2` and there is no oblique asymptote.

Explanation:

`f(x)=frac{5x^2-17x-12}{2x^2-7x-4}`

This problem is a bit tricky because there is a "hole" at `x=4`.

Start by factoring both numerator and denominator.

`f(x)=frac{(5x+3)(x-4)}{(2x+1)(x-4)}`

Note that factor `(x-4)` cancels. This means there is a "hole" or undefined point in the graph at `x=4`.

To find the asymptotes, use the function that results after the factor cancels.

`f(x)=frac{5x+3}{2x+1}`

To find the horizontal asymptote (HA), compare the degree of the numerator to the degree of the denominator.

If the degrees are the same, the HA is `y=` leading coefficient of the numerator divided by the leading coefficient of the denominator.

If the degree of the denominator is greater, the HA is `y=0`.

If the degree of the numerator is greater, there is an oblique asymptote.

`f(x)=frac{color(blue)5x^color(red)1+3}{color(blue)2x^color(red)1+1}`

In this example, both the numerator and denominator have a degree of `color(red)1`. (The degree is the highest exponent in standard form).

The HA is then `y=color(blue)5/color(blue)2`

There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.

To find the vertical asymptote (VA), find the value of `x` where the denominator equals zero. This value of `x` will result in dividing by zero, which is undefined.

`2x+1=0 color(white)(aaa)`Solve for x

`2x=-1`

`x=- 1/2color(white)(aaa)`This is the equation of the VA.