SOLUTION: I would REALLY appreciate it if a tutor could assist me with the following problem: Find all the asymptotes of f(x) = {{{ (2x^2-3x-5) / (x^2-7x+12) }}} I believe there is an

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Question 7446: I would REALLY appreciate it if a tutor could assist me with the following problem:
Find all the asymptotes of f(x) =
I believe there is an oblique and possibly a vertical asymptote.
Thanks so much!!!
Answer by prince_abubu(198) (Show Source):
You can put this solution on YOUR website!
I don't think this has an oblique asymptote. It definitely has a horizontal (<--->) asymptote, though. The way to tell that is to look at the terms with the highest power in both numerator and denominator. We've got the 2x^2 in the numerator and the x^2 in the denominator. You simply just take the numerator and divide by the denominator, and you'll get 2. The horizontal asymptote occurs at y = 2. BTW, the highest powers in both numerator AND denominator HAVE TO MATCH POWERS!

What about the horizontal asymptotes? That's when you have to play around with the denominator. There are "troublemaker" values that you can't plug in to this function because they will make the denominator zero, which you know can't happen. So we'll have to find those troublemaker values that make the denominator zero.

The denominator is . So you ask yourself "what times what equals 12 and their sum or diference must come out to -7?". After a while, you'll come up with -3 and -4. So then you write (x - 3)(x - 4). The values 3 and 4 will make the denominator zero, so you CAN'T plug those into your function. Graphically, this gives you the vertical asymptotes x=3 and x=4.

I don't remember the rules for having an oblique asymptote, though. If I can remember correctly, an oblique happens when the term with the highest power in the numerator is of a higher power than the denominator's highest power term.