Question

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

Answer 1

find c?

Answer 2

tell me if 1/x^2-c is a fraction form?

Answer 3

Yes it is in fraction form, 1 on top and x^2-c on the bottom

Answer 4

this is calculus y do u want to find c

Answer 5

I don't know... It confuses me... This is more of an algebra question than calculus.. and I sucked at algebra....

Answer 6

i cant find c with f(x) thingie

Answer 7

Okay.. Thanks anyway :)

Answer 8

c=x^2-1/y there u go

Answer 9

If f has a vertical asymptote at x=3, then that means the function shouldn't exist at x=3.
The function doesn't exist when the fraction doesn't exist. The fraction doesn't exist when the denominator is 0 (when the bottom is 0).

When is x^2-c equal to 0?
And since the vertical asymptote is at x=3
then we no 3^2-c equal to 0 will give us a value of c that makes the function undefined at x=3.

Answer 10

know not no (darn it)

Answer 11

so c=9?

Answer 12

Yep! That would give us f(x)=1/(x^2-9)
As you can see this function does not exist at x=3 (or x=-3 <---but whatever :p)

Answer 13

Awesome!! Thanks a million!!! Do you think you could help me with this one too? Find the equation of the slant asymptote for the graph of y=\[y=\frac{ 3x ^{2}+2x-3 }{ x-1 }\] @myininaya

Answer 14

All you really need to do is divide.

Answer 15

I know I have to turn it into a division problem.. But after that I'm not sure... Because the long division with Xs confuses me..

Answer 16

If you can do the division correctly and if the numerator is 1 degree higher than the numerator, then the slant asymptote will be the quotient.