Question

Consider the function f(x) = (x^2-9)/(x-4) . Determine all asymptotes of this function including horizontal, vertical, and oblique (slant).

Answer 1

the degree of the numerator is 2 and the degree of the denominator is 1, so there is no vertical asymptote, but there is a slant asymptote because the degree of the numerator is larger by 1

Answer 2

the vertical asymptote is easiest, it is where the function is undefined, where the denominator is 0
you can solve that with your eyeball right?

Answer 3

you find the "slant asymptote" by division, i.e. divide \(x^2-9\) by \(x-4\) you can use synthetic division if you know it, otherwise you have to use polynomial long division, which is harder

Answer 4

how do you feel about synthetic division?

Answer 5

I think i'm fine with synthetic division, thanks a lot!

Answer 6

when you do it you can ignore the remainder, because it is only the line that you need

Answer 7

yw

Answer 8

Sorry, one more question when I did synthetic division I got the numbers 1 4 25. Is that right..?

Answer 9

And how would you find horizontal asymptote?

Answer 10

ok there is NO horizontal asymptote because the degree of the numerator is larger than the degree of the denominator
that is why you are finding the slant asymptote

Answer 11

1 4 25 is wrong actually, it should be 1 4 7 but in fact is doesn't matter because this means
\[\frac{x^2-9}{x-4}=x+4+\frac{7}{x-4}\] but the remainder is unimportant

Answer 12

And there is also no vertical asymptote just by looking.

Answer 13

the line \(y=x+4\) is your slant asymptote

Answer 14

forget the remainder part, that is just the remainder

Answer 15

oh yes indeed there is a vertical asymptote
you get it by setting the denominator equal to zero and solve

Answer 16

\[x-4=0\iff x=4\] and so your vertical asymptote is the vertical line \(x=4\)

Answer 17

you good with this?

Answer 18

Yeah I actually think I am, thanks again.