How do you find all the asymptotes for function #(x^2+1)/(x^2-9)#?

1 Answer
Jul 31, 2015

Answer:

This function has vertical asymptotes at #x=\pm 3# and a horizontal asymptote at #y=1#.

Explanation:

For the horizontal asymptote, note that the rational function #f(x)=(x^2+1)/(x^2-9)# has numerator and denominator polynomials of the same degree 2. Therefore, its graph will have a horizontal asymptote at the value of #y# that is the ratio of the coefficients of the highest powers of #x# in the numerator and denominator, which is #1/1=1#; #y=1# is the (unique) horizontal asymptote.

For the vertical asymptotes, the best thing to do initially in general is to factor the numerator and denominator as much as possible and cancel any common factors. In this case, the numerator cannot be factored further over the real numbers and the denominator has no common factors with the numerator: #f(x)=(x^2+1)/((x-3)(x+3))#.

Because nothing cancels, the vertical asymptotes will occur at the values of #x# where the denominator is zero, which are #x=\pm 3#.

Here's a picture of the graph of this function:

graph{(x^2+1)/(x^2-9) [-20, 20, -10, 10]}