Question

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

Answer 1

Vertical asymptote: `x=2`
No horizontal asymptote
Oblique asymptote: `y=2x+9`

Explanation:

Vertical asymptotes: we must look for points in which the function is not defined. In this case, the point is `x=2`, since it causes the denominator to vanish:

`f(2)=\frac{2*2^2+5*2}{2-2}=\frac{18}{0}`

which is undefined.

Horizontal asymptotes: we can look for horizontal asymptotes, since the domain is unbounded. If they exist, they are given by `lim_{x\to\pm\infty}f(x)`:

`\lim_{x\to\pm\infty}\frac{2x^2+5x}{x-2}=\frac{x(2x+5)}{x(1-\frac{2}{x})} = \frac{2x+5}{1-\frac{2}{x}}\to\frac{\pm\infty}1 = \pm \infty`

Oblique asymptotes: we can look for oblique asymptotes, since the function diverges. Here's the theory: if `y=f(x)` approaches some line `y=mx+q` as `x\to\pm\infty`, then we have

`f(x)-(mx+q) \to 0 \iff f(x)\to mx+q`

If we divide both sides by `x`, we have

`f(x)/(x) \to (mx)/x+q/x \to m`

since `q/x\to 0` because `x\to\pm\infty`

So, if `f(x)/x` has a finite limit, that would be the slope of the line it is approaching. Once `m` is known, we simply have

`f(x)\to mx+q \iff f(x)-mx \to q`.

Let's do it!

`m = \lim_{x\to\pm\infty}f(x)/(x)= \lim_{x\to\pm\infty}\frac{2x^2+5x}{x^2-2x}=2`

`q = \lim_{x\to\pm\infty}f(x)-mx = \lim_{x\to\pm\infty}\frac{2x^2+5x}{x-2}-2x`
`=\lim_{x\to\pm\infty}\frac{9x}{x-2}=9`