## anonymous 3 years ago Determine all horizontal, slant, and vertical asymptotes. For each vertical asymptote, determine whether f(x) -> infinity sign of f(x) -> negative infinity sign on either side of the asymptote.

1. anonymous

f(x) = x^2/4 - x^2

2. anonymous

is it $\frac{x^2}{4-x^2}$?

3. anonymous

yes

4. anonymous

ok so for the vertical asymptotes, set the denominator equal to zero and solve for $$x$$

5. anonymous

ok

6. anonymous

you get $4-x^2=0$ or $(2-x)(2+x)=0$ and the solutions are $$x=-2$$ or $$x=2$$ those are your vertical asymptotes (there are two of them)

7. anonymous

ok

8. anonymous

for the horizontal asymptote, note that the numerator and denominator have the same degree (both are degree 2) so it is the ratio of the leading coefficients

9. anonymous

okay

10. anonymous

the leading coefficient of $$x^2$$ is 1 and the leading coefficient of $$4-x^2$$ is $$-1$$ therefore the horizontal asymptote is $$y=-1$$

11. anonymous

that is in the case where the degrees are the same. there is no slant asymptote for there to be a slant asymptote , the degree of the numerator would have to be one more than the degree of the denominator

12. anonymous

okay thx

13. anonymous

yw