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

May 29, 2016

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 ${x}^{2} - 2 x - 8 = 0$
By factorising: $\left(x - 4\right) \left(x + 2\right) = 0$
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 $x = 3.9 \mathmr{and} 4.1$ 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, ${\lim}_{x \rightarrow {4}^{-}} f \left(x\right)$ 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 $f \left(x\right)$ 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 $\frac{1}{x} ^ 2$ will affect the value of f(x) significantly. But in this case, the asymptote is $f \left(x\right) = 0$ (or $y = 0$) since if $x$ tends to infinity then $f \left(x\right)$ will get increasingly smaller.

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