Building Functions from Asymptotes

[Lizardjuice7]

Homework Statement

Use the given information about the asymptotes of a function to write an equation for a rational function. You may leave your function in "factored" form.

Homework Equations

1. Vertical Asymptote at x=3 and x=-2

2. Vertical Asymptote at x=5, Horizontal Asymptote at y=3

3. Hole at (2,3), Horizontal Asymptote at y=0

4. Horizontal Asymptote at y=0, Vertical Asymptote at x=6, Goes through the point (3,-2)

5. Vertical Asymptote at x=2, Oblique Asymptote with Equation y=3x-5

The Attempt at a Solution

I know the basics for finding horizontal, vertical, and oblique asymptotes when you have the equation. I'm just having a hard time going backwards.

1. f(x)=1/(x-3)(x+2)

2. f(x)=3x/(x-5)

3. I know the denominator has to be a higher degree than the numerator, bu the "hole" in the graph through me for a loop. No idea where to start.

4. I'm having a similar problem, I don't know how to find a point that goes through the asymptote.

5. f(x)=3x²-11x+10/(x-2)

I got that by setting the denominator equal to zero to find '(x-2)'. I then multiplied the oblique asymptote 'y=3x-5' by '(x-2)' to get the numerator. I think i did something wrong though.

Any help would be greatly appreciated!

Thanks
Lizardjuice7

 

[Dick]

None these problems necessarily determines a unique function, you just have to find one that works. 1 and 2 look good. A 'hole in the graph' is referring to a function like (x-2)/(x-2). It's equal to 1 everywhere except at x=2, where it's undefined. That's the 'hole in the graph'. For 4 the point (-3,2) really supposed to be on the asymptote or on the graph? It's not on either asymptote. For 5 the equation should be (3x-5) times something that goes to 1 as x->infinity, right? And that something should have a vertical asymptote at x=2.