# Math-precalc

posted by
**Pamela**
.

Is it possible to find a rational function that has x-intercepts (-2,0) and (2,0), but has vertical asymptote x=1 and horizontal asymptote of y=0?

The horizontal asymptote and the x-intercepts parts stump me. If you can't reach y=0, then how can you get the x-intercepts?

And I don't know how to start finding a rational function that has asymptotes x=0 and y=x. It seems like I only have enough space to place a function in Q-2, Q-4.

Well, you can reach y = 0 at x = ±2, but you'll also reach it in the limit x to ± infinity.

You just write down the polynomila that has zeros ar ±2:

x^2 - 4

you divide this by a function whiich has a zero at x = 1: x - 1. You then get:

(x^2-4)/(x-1)

But this function does not go to zero if x goes to infinity. So, you divide this by a polynomial that has no zeroes to make the degree of the denominator larger than the degree of the numerator. You can, e.g. take that polynomial to be x^2 +1. Then you get:

(x^2-4)/[(x-1)(x^2+1)]

And I don't know how to start finding a rational function that has asymptotes x=0 and y=x. It seems like I only have enough space to place a function in Q-2, Q-4."

You can take:

1/x + x