## haleyelizabeth2017 one year ago 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.

1. SolomonZelman

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

2. jim_thompson5910

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

3. SolomonZelman

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

4. haleyelizabeth2017

C is equal to 9

5. SolomonZelman

Yes

6. haleyelizabeth2017

so $\pm 9$ are the vertical asymptotes?

7. SolomonZelman

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

8. jim_thompson5910

no, you just found c = 9 9 isn't an asymptote

9. haleyelizabeth2017

oh, whoops

10. SolomonZelman

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

11. haleyelizabeth2017

Wait...c has two values?

12. jim_thompson5910

nope, just one and it's c = 9 SolomonZelman is asking a slightly different related question

13. haleyelizabeth2017

Okay...

14. SolomonZelman

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*.

15. SolomonZelman

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.

16. haleyelizabeth2017

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

17. SolomonZelman

any questions?

18. haleyelizabeth2017

Nope, thank you very much

19. SolomonZelman

Yw