for h(x) = (-5x^2) / ( x^2 -4)

is x= 2 and x= -2 an symtope for this? for vertical asym

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- Nov 23rd 2008, 09:07 AMrj2001asymtope help
for h(x) = (-5x^2) / ( x^2 -4)

is x= 2 and x= -2 an symtope for this? for vertical asym - Nov 23rd 2008, 09:28 AMo_O
Yep. (Yes)

To prove that they are, take any one-sided limit for each of them.

- $\displaystyle \lim_{x \to 2^-} f(x) = \cdots$ ... or ... $\displaystyle \lim_{x \to 2^+} f(x) = \cdots$
- $\displaystyle \lim_{x \to -2^-} f(x) = \cdots$ ...or... $\displaystyle \lim_{x \to -2^+} f(x) = \cdots$

- $\displaystyle \lim_{x \to 2^-} f(x) = \cdots$ ... or ... $\displaystyle \lim_{x \to 2^+} f(x) = \cdots$
- Nov 23rd 2008, 10:07 AMrj2001
should the graph look like this?

y in the -5 area horizontal asymtope