Question

a rational function with a removable discontinuity at x=-3, a vertical asymptote at x=3, a horizontal asymptote at y=0 and a y-intercept at (0,-1/3)
can someone translate this to english?

Answer 1

There would be an unallowable value in the denominator at x = 3 because there would then be a "0" in the denominator, thus also giving a vertical asymptote at x = 3. A removable discontinuity at x = -3 could occur if you have a factor of (x + 3) in both the numerator and the denominator. Y-intercept at that value is for when x = 0. We could try to create such a reprentative function for you.

Answer 2

The last part, horizontal asymptote at y = 0 is for x going off to +-infinity, the function would have to be approaching "0".

Answer 3

A very simple function that fits this scenario is:\[y = \frac{ x+3 }{ x ^{2}-9 }\] because you can factor out "x + 3" from the numerator and the denominator so the discontinuity at x = -3 is removed.

Answer 4

Also, as x goes to +-infinity, the function value approaches 0. Y-intercept value and vertical asymptote are satisfied.

Answer 5

Making sense to you now?

Answer 6

yes! I couldnt translate it for myself, thank you:)