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.
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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.
Thank you. How do you find the
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.
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There IS a hole, however.
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).