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r(x)=x^3+3x^2/x^2-4

2006-11-18 08:17:17 · 5 answers · asked by jere 1 in Science & Mathematics Mathematics

5 answers

There are no horizontal asymptotes because the degree of the numerator is greater than the denominator.

2006-11-18 08:24:09 · answer #1 · answered by hayharbr 7 · 0 0

Vertical asymptotes are easy enough, just find out what x puts 0 in the denominator.

2006-11-18 16:20:50 · answer #2 · answered by Jacob P 2 · 0 1

(x^3+3x^2)/x^2-4
x^2(x + 3)/(x + 2)(x - 2)

vertical asymptotes
x = -2, 2

horizontal asymptotes
none

2006-11-18 16:58:51 · answer #3 · answered by trackstarr59 3 · 0 0

VA: x+2 and x= -2
HA: none

2006-11-18 16:22:27 · answer #4 · answered by      7 · 0 0

V.A : -2 , 2
H.A. : none

2006-11-18 16:42:43 · answer #5 · answered by Gardenia 6 · 0 0

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