How do you find the asymptotes for #h(x)= (x^2-4)/(x)#?
1 Answer
Jun 1, 2016
oblique asymptote y = x
Explanation:
The first step is to divide out.
#rArr(x^2-4)/x=(x^2/x-4/x)=x-4/x# thus h(x) simplifies to
#h(x)=x-4/x# as
#xto+-oo,4/xto0" and " h(x)tox#
#rArry=x" is the only asymptote"#
graph{(x^2-4)/x [-10, 10, -5, 5]}
