Question

will someone please explain how to find the vertical asymptotes of:
f(x)= (2x^3-15x^2-29x+102)/(x^3-5x^2-136x+140)

Answer 1

Easier to read:

\(f(x)= (2x^3-15x^2-29x+102)/(x^3-5x^2-136x+140)\)

Answer 2

thank you! do you know how to find the vertical asymptotes of it?

Answer 3

Find the roots of the denominator.

If they do not make the numerator zero, then you have an vertical asymptote.

Answer 4

tip2 : try x=1

Answer 5

denominator would equal 2.. not 0. so now that i know there is an asymptote, how do i find it>

Answer 6

What !!!!

Try again: 1-5-136+140 = ?

Answer 7

oh hahaha whoops. i for some reason i read 136 as 134. so there is not a vertical asymptote?

Answer 8

Now, does x=1 make the numerator zero?

Answer 9

yes

Answer 10

2-15-29+102 = ?

Answer 11

60

Answer 12

So numerator is not zero when x=1, so x=1 is one of the vertical asymptotes.

Find the other ones by the same method.
Your denominator should simplify, as you can factorize (x-1)

Answer 13

are there going to be 3 vertical asymptotes?

Answer 14

3 is the maximum you can obtain with a cubic polynomial as denominator.

Actually, there will be 3 in your case.

Answer 15

do i just use guess and check to find the rest?

Answer 16

No, write \(x^3-5x^2-136x+140\) as \((x-1)(x-a)(x-b)\)

a and b will be the new values of vertical asymptotes.

Do you know polynomial division?