How do you find any asymptotes of #h(x)=(x-5)/(x^2+2x-4)#?

1 Answer
Dec 11, 2016

Answer:

By (a) seeing if the denominator is zero for any value(s) of #x#, and by examining the behaviour for large positive and negative values of #x#.

Explanation:

The denominator is zero where #x=-1+-sqrt 5# so there #h(x)# has vertical asympototes at #x=-1+sqrt 5# and #x=-1=sqrt 5#.

For large #x# positive or negative #h(x)# approximates to #1/x# (#h(x)to 0# as #x to +- oo#), as you can see by ignoring the #-5# and the #2x-4# for large #x# magnitudes.

So the #x# axis is another asymptote.