Question

How do you find vertical, horizontal and oblique asymptotes for `(3x^2+2x-5)/(x-4)`?

Answer 1

The vertical asymptote is `x=4`
The oblique asymptote is `y=3x+14`
No horizontal asymptote

Explanation:

Let `f(x)=(3x^2+2x-5)/(x-4)`

The domain of `f(x)` is `D_f(x)=RR-{4}`

As we cannot divide by `0`, `=>`, `x!=4`

The vertical asymptote is `x=4`

As the degree of the numerator is `>` than the degree of the denominator, there is an oblique asymptote.

Let's do a long division

`color(white)(aaaa)` `3x^2+2x-5` `color(white)(aaaaaa)` `∣` `color(blue)(x-4)`

`color(white)(aaaa)` `3x^2-12x` `color(white)(aaaaaaaaa)` `∣` `color(red)(3x+14)`

`color(white)(aaaaaa)` `0+14x-5`

`color(white)(aaaaaaaa)` `+14x-56`

`color(white)(aaaaaaaaaa)` `+0+51`

Therefore,

`f(x)=(3x+14)+51/(x-4)`

Now, we calculate the limits

`lim_(x->-oo)(f(x)-(3x+14))=lim_(x->-oo)51/x=0^-`

`lim_(x->+oo)(f(x)-(3x+14))=lim_(x->+oo)51/x=0^+`

The oblique asymptote is `y=3x+14`

graph{(y-(3x^2+2x-5)/(x-4))(y-3x-14)=0 [-106, 160.9, -31.5, 102.1]}