pre-calc

what are the asymtopes of

2x/x-3

asked by caitlyn
  1. First make sure the expression is correctly transcribed:
    2x/x-3
    does not equal
    2x/(x-3).
    I suspect it's the latter you need.

    Because of the lack of typesetting capabilities, parentheses are required around the numerator and the denominator.


    Hints for asymptotes and domains:

    1. All polynomials have a domain of R, or (-∞,∞).
    2. Horizontal asymptotes are finite values to which the function approaches when x→±∞.
    3. Vertical asymptotes for a rational function (i.e. polynomials in both numerator and denominator) occur when the denominator equals zero.
    4. The domain of a rational function is therefore (-∞,∞)-points at which vertical asymptotes occur.

    Post your answers for checking if you wish.

    posted by MathMate

Respond to this Question

First Name

Your Response

Similar Questions

  1. Check my precalc

    Find the verticle, horizontal, and oblique asymtopes if any. #1 (x^2 +6x+5)/2x^2 +7x+5 y= (1/2) and x= -5/2 and -1 #2 (8x^2 +26x-7)/4x-1 y=2x+7 and x=1/4 #3. (x^4-16)/(x^2-2x) x= 0 and 2 y= x^2 +2x+4 thanks!
  2. Math

    for function f(x)=(x)/x^2-4. 1. State the vertical and horizontal asymtopes. 2. Graph this function.

More Similar Questions