# How do you find the asymptotes for f(x) = (x^2-25)/(x^2+5x)?

Jan 13, 2016

The asymptote is x=0

#### Explanation:

The reason for that is this:

$\left({x}^{2} - 25\right) \div \left({x}^{2} + 5 x\right)$ can be simplified to

$\left(x - 5\right) \left(x + 5\right) \div x \left(x + 5\right)$

The (x+5) will divide out, leaving the equation as $\left(x - 5\right) \div x$. Now, the definition of an asymptote is "a line that continually approaches a given curve but does not meet it at any finite distance" (Oxford Dictionaries). What that translates to is a line that will not ever touch a certain value. We can tell what that value is by determining the number which would make the denominator of our equation zero. And if a denominator is 0, you are dividing by zero, which is" illegal" in math laws.

So, whatever number makes the denominator of $\left(x - 5\right) \div x$ equal zero, that is the asymptote. In this case, the value that makes the denominator zero is zero, so it's fairly easy.

Other information that might be useful is that there is a hole at -5 and there is an x-intercept at (5,0).
Hope this helped!