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.

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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.

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

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Other answers:

C is equal to 9
Yes
so \[\pm 9\] are the vertical asymptotes?
No, the vertical asymptotes are: *±√9*
no, you just found c = 9 9 isn't an asymptote
oh, whoops
Yes, so your asymptotes are? (just to verify)
Wait...c has two values?
nope, just one and it's c = 9 SolomonZelman is asking a slightly different related question
Okay...
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*.
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.
Sorry, I was confused before, so I had to clarify lol
any questions?
Nope, thank you very much
Yw

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