Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)=(x+2)/sqrt(6x^2+5x+4)`?

Answer 1

`f(x)` has horizontal asymptotes `y = sqrt(6)/6` and `y = -sqrt(6)/6`

Explanation:

Note that:

`6x^2+5x+4`

is in standard quadratic form with `a=6`, `b=5` and `c=4`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (color(blue)(5))^2-4(color(blue)(6))(color(blue)(4)) = 25-96 = -71`

Since `Delta < 0`, this quadratic has no real zeros and since its leading coefficient is positive, it always takes positive values.

So the denominator `sqrt(6x^2+5x+4)` of `f(x)` is always well defined and non-zero.

So `f(x)` has no vertical asymptotes.

Note that:

`lim_(x->+oo) f(x) = lim_(x->+oo) (x+2)/sqrt(6x^2+5x+4)`

`color(white)(lim_(x->+oo) f(x)) = lim_(x->+oo) (1+2/x)/sqrt(6+5/x+4/x^2)`

`color(white)(lim_(x->+oo) f(x)) = 1/sqrt(6)`

`color(white)(lim_(x->+oo) f(x)) = sqrt(6)/6`

`lim_(x->-oo) f(x) = lim_(x->-oo) (x+2)/sqrt(6x^2+5x+4)`

`color(white)(lim_(x->-oo) f(x)) = lim_(x->+oo) (-x+2)/sqrt(6x^2-5x+4`

`color(white)(lim_(x->-oo) f(x)) = lim_(x->+oo) (-1+2/x)/sqrt(6-5/x+4/x^2)`

`color(white)(lim_(x->-oo) f(x)) = -1/sqrt(6)`

`color(white)(lim_(x->-oo) f(x)) = -sqrt(6)/6`

So `f(x)` has horizontal asymptotes `y = +-sqrt(6)/6`

That leaves no opportunity for slant asymptotes.

graph{(x+2)/sqrt(6x^2+5x+4) [-22, 22, -2.52, 2.48]}