How do I identify the horizontal asymptote of #f(x) = (7x+1)/(2x-9)#?

1 Answer
Mar 9, 2018

We have a horizontal asymptote #y=3.5#

Explanation:

As the degree of polynomial in the numerator is equal to the degree of polynomial in the denominator, there is indeed a horizontal asymptote. We can find this by dividing each term in numerator and denominator by this highest degree and find limit as #x->oo#. The process is shown below:

Now #lim_(x->oo)(7x+1)/(2x-9)#

= #lim_(x->oo)(7+1/x)/(2-9/x)#

= #7/2#

Hence, we have a horizontal asymptote #y=7/2# or #y=3.5#

graph{(y-(7x+1)/(2x-9))(y-3.5)=0 [-40.42, 39.58, -17.76, 22.24]}