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

Question

Lincoln Anderson · Accepted Answer

We have a horizontal asymptote #y=3.5#

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]}

Elias Ackerman · Answer

undefined To identify the horizontal asymptote of $f(x) = \frac{7x+1}{2x-9}$, compare the degrees of the polynomials in the numerator and the denominator:

- The degree of the numerator (the highest power of $x$ in the numerator) is 1.
- The degree of the denominator (the highest power of $x$ in the denominator) is 1.

When the degrees of the numerator and denominator are equal, the horizontal asymptote is found by dividing the leading coefficients of the numerator and the denominator.

The leading coefficient of the numerator is 7, and the leading coefficient of the denominator is 2. Therefore, the horizontal asymptote is $y = \frac{7}{2}$.