Question

help!! find the asymptotes. i really have no clue how to do this. x^3-1/x^2-3x+2

Answer 1

To find the asymptotes, you need to let the denominator equal to zero and then solve for x(for this particular problem you'll need to do some factoring to get your solution).

Answer 2

i got x=sqrt(8/3). ugh! i need to find the vertical, horizontal, and slant. I think there are no horizontal asymptotes. is the vertical asymptote sqrt8/3?

Answer 3

or x=1, x=2

Answer 4

Yeah, there are no horizontal because the x in the numerator is being raised to a greater degree(3) than the x on the bottom(2). For the vertical, your second answer is right :)

Answer 5

do you know about the slant?

Answer 6

Problems all about.

1) Bad notation

x^3-1/x^2-3x+2 this is NOT what you intend. This is (x^3-1)/(x^2-3x+2). The parentheses are NOT optional.

2) Bad form.

If you want a rational function to be represented on a graph, you should show that.

y = (x^3-1)/(x^2-3x+2) The "y = " is NOT optional.

Always be on the lookout for simplification.

y = (x^3-1)/(x^2-3x+2) = (x^2 + x + 1)/(x-2) for \(x \ne 1\) This leaves a hole in the Domain, and entirely simplifes the effort.

Reduce the degree of the numerator by long division.

y = (x^2 + x + 1)/(x-2) = x + 3 + 7/(x-2)

This clearly exposes the oblique asymptote, y = x + 3
The vertical asymptote is also exposed, x = 2

Are we close?