Question

What is the vertical asymptote, horizontal asymptote and the removable discontinuities of the following 4 problems?

Answer 1

First, for vertical asymptote, you find the x-intercept of the denominator.
http://www.purplemath.com/modules/asymtote.htm

Answer 2

For the horizontal asymptote, apparently you have to make a chart http://www.purplemath.com/modules/asymtote2.htm

Answer 3

Whereas you sub in extreme values for both the positives and negatives, to represent positive and negative infinity.

Answer 4

For removable discontinuities, I know I did really bad in that part of the unit.
http://oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/removable.html

Answer 5

Maybe @Campbell can help there?

Answer 6

(1) vertical asymptote at x = 0... since 3x cannot = 0

use polynomial division for oblique asymptotes



the oblique asymptote is y = 1/3x - 4

(2) vertical asymptote at x = -4 and a horizontal asymptote at y = 0

(3) no real idea... you need to find the roots of the cubics

but a horizontal asymptote exists at y = 2 from polynomial division

(4) factorise the numerator and denominator


y = [(x +5)(x-1)]/[(x +3)(x + 5)]

vertical asymptote at x = -3, point of discontinuity at x = -5

horizontal asymptote at x = 1 ( found by polynomial division)