Question

How do you find the asymptotes for `f(x) = 1 /(x^2-2x-8)`?

Answer 1

See below.

Explanation:

In this particular case, only two things cause asymptotic behaviour: if the denominator is zero, or as `x` tends to infinity.

Let's study these individually:

If the denominator is zero:
Then `x^2-2x-8=0`
By factorising: `(x-4)(x+2)=0`
Finding solutions: `x=4` or `x=-2`

Therefore, the above lines will cause asymptotic behaviour as the curve approaches. You can substitute values of x such as `x=3.9 or 4.1` to determine whether the curve approaches positive or negative infinity from these points (since limits are usually defined as having a positive and negative side, for example, `lim_(xrarr4^-)f(x)` which would be determined by checking `x=3.9`)

As `x` tends to infinity:
As this occurs, only the term which the largest order will matter (order refers to the power or exponent value). We consider both the top and bottom of the fraction of `f(x)` separately. The highest order of the top is just `1` (`x^0`) and on the bottom this is `x^2`. So as `x` tends to infinity, only the `1/x^2` will affect the value of f(x) significantly. But in this case, the asymptote is `f(x)=0` (or `y=0`) since if `x` tends to infinity then `f(x)` will get increasingly smaller.

So the three asymptote said are `x=4`, `x=-2` and `y=0`.