Question

The graph of f(x)=1/(x^2-c) has a vertical asymptote at x=3. Find c.

I know the vertical asymptote deals only with the denominator, which in this case, is x^2-c.

Answer 1

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-c} }\)

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{(x-\sqrt{c})(x+\sqrt{c})} }\)

Answer 2

`The graph of f(x)=1/(x^2-c) has a vertical asymptote at x=3` means f(3) is undefined and the denominator `x^2-c` is equal to 0 when x = 3


x^2-c = 3^2 - c = 9 - c = 0


solve 9-c = 0 for c

Answer 3

This function will have two vertical asymptotes:
*[1]* \(\large\color{black}{ \displaystyle x=\sqrt{c}}\)
*[2]* \(\large\color{black}{ \displaystyle x=-\sqrt{c}}\)

Answer 4

C is equal to 9

Answer 5

Yes

Answer 6

so \[\pm 9\] are the vertical asymptotes?

Answer 7

No, the vertical asymptotes are: *±√9*

Answer 8

no, you just found c = 9

9 isn't an asymptote

Answer 9

oh, whoops

Answer 10

Yes, so your asymptotes are? (just to verify)

Answer 11

Wait...c has two values?

Answer 12

nope, just one and it's c = 9

SolomonZelman is asking a slightly different related question

Answer 13

Okay...

Answer 14

Yeah, C is one value and it is 9. (we said that before)

\(\large\color{black}{ \displaystyle f(x)=\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)} }\)

the function is undefined (because of the denominator) at *x=-3* and *x=3*.

Answer 15

but, the answer to your question (to find the number C, such that there is going to be a vertical asymptote at x=3), that is C=9.

Answer 16

Sorry, I was confused before, so I had to clarify lol

Answer 17

any questions?

Answer 18

Nope, thank you very much

Answer 19

Yw