# 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?

##### 1 Answer

#### 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

Given that we want our function to cross the

Given that we want it to touch the

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

#(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

#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]}