Question

For the following function f(x)=(x^2-4)/x^2-x-6, find the horizontal asymptote, oblique asymptote and hole in the graph (if any). Please show steps.

Answer 1

Do you remember the 3 cases? For this problem, we see that the numerator and the denominator both have a degree of 2 (they both have 2 as the biggest exponent). In that case, there is a horizontal asymptote y = a. To find that number a, you take the quotient of the coefficients of the highest powers. 1x^2 (numerator) and 1x^2 (denominator) means that a = 1/1=1. So, there is a horizontal asymptote y = 1.

Answer 2

Thank you. How do you find the
oblique asymptote?

Answer 3

There will only be an oblique (slant) asymptote if the degree of the numerator is 1 greater than the degree of the denominator, so there isn't one.

Answer 4

There IS a hole, however.

Answer 5

You have to factor the top and the bottom and see if things cancel to see if there's a hole... The hole happens at x = -2, because x + 2 cancels out. The y-value of the hole is found by plugging -2 into what's left after cancelling: ((-2) - 2)/(-2 - 3) = 4/5. So, there's a hole at (-2, 4/5).

Answer 6

Thank you. You've explained that so well.