rational function w/ vertical asympt's at x = -2, x = 3, x-int at 0, hor. asympt. y=2

[noah2793]

[h=1]Determine a rational function that has vertical asymptotes at x = -2 and x = 3 and, an x-intercept at 0, and a horizontal asymptote of 2.?[/h]
I have gotten to 2x^3/x^3-x^2-6x but I can't figure out to get it so the x intercept is at 0.

 

[Deleted member 4993]

Determine a rational function that has vertical asymptotes at x = -2 and x = 3 and, an x-intercept at 0, and a horizontal asymptote of 2.?


I have gotten to 2x^3/x^3-x^2-6x but I can't figure out to get it so the x intercept is at 0.

Can you please explain how did you arrive at that answer? Does not quite make sense to me.

 

[Dr.Peterson]

Determine a rational function that has vertical asymptotes at x = -2 and x = 3 and, an x-intercept at 0, and a horizontal asymptote of 2.?


I have gotten to 2x^3/x^3-x^2-6x but I can't figure out to get it so the x intercept is at 0.

You mean 2x^3/(x^3-x^2-6x), right?

The function you wrote is undefined at x=0. What can you do to fix that? One small change will yield the correct answer; most of your work is good.

 

[topsquark]

Determine a rational function that has vertical asymptotes at x = -2 and x = 3 and, an x-intercept at 0, and a horizontal asymptote of 2.?


I have gotten to 2x^3/x^3-x^2-6x but I can't figure out to get it so the x intercept is at 0.

What you are trying to do in the red equation is \(\displaystyle 2x^3 / (x^3 - x^2 - 6x)\). Parenthesis are important.

What you are reaching for is something along the lines of \(\displaystyle \dfrac{2x^3}{x ( x + 2)(x - 3)}\). But what is the value of this when x = 0? How can you fix this problem? (Hint: How many x's do you need in the denominator? Can you cancel an x somewhere?)

-Dan

Dr.Peterson beat me to it!