Question

tell me how to do this? stuck!


Identify the vertical asymptotes of f(x) = quantity x plus 6 over quantity x squared minus 9x plus 18

Answer 1

if the top and bottom have no like factors, the zero of the bottom creates a vertical asymptote

Answer 2

\[f(x)=\frac{x+6}{x^2-9x+18}\]

Answer 3

you can do as @amistre64 said

Answer 4

for example:\[\frac{(x-k)}{(x-k)(x-c)}\]
the zero for (x-c) has no common factor above so it creates a vertical asymptote at x=c

Answer 5

I am still lost. Mind using my question for example for explaining it a bit?

Answer 6

factor your denominator

Answer 7

Are there any factors of the denominator which can equal 0? If so, the point at which they equal 0 is a vertical asymptote.

For example
\[\frac{1}{x-1}\] has a vertical asymptote at \(x = 1\) because at that point you would be dividing by 0

Answer 8

(x-6)(x-3)

Answer 9

good, now is there an (x-6) or an (x-3) on the top?

Answer 10

nope

Answer 11

then there is not way to cancel out one of the factors, it will have a vertical asymptote at both zeros of the denominator

Answer 12

(to clarify my comment: I'm talking about after canceling any common factors from numerator and denominator)

Answer 13

x-6 = 0, when x=?

x-3 = 0, when x=?

Answer 14

x=6 & x=3

Answer 15

correct, so those are the vertical asymptotes

Answer 16

thank you!

Answer 17

youre welcome

Answer 18

side note: in control system theory, where you do a lot of this examination of ratios of polynomials, the spots where the numerator = 0 are called zeros (big surprise, as the fraction then equals 0), and the spots where the denominator = 0 are called poles (because the value of the function sticks up like the pole in the middle of the circus tent).