## onegirl 2 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. onegirl

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

2. satellite73

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

3. onegirl

yes

4. satellite73

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

5. onegirl

ok

6. satellite73

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. onegirl

ok

8. satellite73

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. onegirl

okay

10. satellite73

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. satellite73

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. onegirl

okay thx

13. satellite73

yw