Question

How do you create a rational function that includes the following: crosses x-axis at 4, Touches the x-axis at -3, One vertical asymptote at x=1 and another at x=6, one horizontal asymptote at y= -3?

Answer 1

`f(x) = (-3x^3-6x^2+45x+108)/(x^3-8x^2+13x-6)`

Explanation:

The denominator of the rational function is primarily responsible for vertical asymptotes: If the denominator is zero and the numerator non-zero then there will be a vertical asymptote.

So given that we want vertical asymptotes at `x=1` and `x=6`, let's consider a denominator `(x-1)(x-6) = x^2-7x+6`

Given that we want our function to cross the `x` axis at `x=4`, we need a factor `(x-4)` in the numerator.

Given that we want it to touch the `x` axis at `x=-3` we need factors `(x+3)(x+3)`

So the numerator needs to be a multiple of:

`(x-4)(x+3)(x+3) = (x-4)(x^2+6x+9) = x^3+2x^2-15x-36`

So far, our function looks like:

`(x^3+2x^2-15x-36)/(x^2-7x+6)`

This has all of the attributes that we want except the horizontal asymptote.

To get a horizontal asymptote at a non-zero value of `y`, we need the numerator and denominator to have the same degree. To avoid adding any extra vertical asymptotes, we can duplicate one of the existing linear factors in the denominator, say `(x-1)` to get:

`(x^3+2x^2-15x-36)/((x-1)(x^2-7x+6)) = (x^3+2x^2-15x-36)/(x^3-8x^2+13x-6)`

This has horizontal asymptote `y=1`. To change it to `y=-3` just multiply the whole function by `-3` to get:

`f(x) = (-3(x^3+2x^2-15x-36))/(x^3-8x^2+13x-6)`

`color(white)(f(x)) = (-3x^3-6x^2+45x+108)/(x^3-8x^2+13x-6)`

graph{(-3x^3-6x^2+45x+108)/(x^3-8x^2+13x-6) [-25.05, 25.07, -12, 12]}