g(x) = 3x^2 + 2x-1 / x^2 -4
There are vertical asymptotes at:
a. x = 2 and x = -2
b. x = 4 only
There is a horizontal asymptote at:
a. y = -1
b. y = 3
2007-12-16 18:36:16 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
g (x) = y = (3x + 1) (x - 1) / (x - 2)(x + 2)
Gives vertical asymptotes at x = ± 2
OPTION a
g(x) = 3 + (2x + 1) / (x² - 4)
Gives horizontal asymptote y = 3
OPTION b
2007-12-16 21:52:36 · answer #1 · answered by Como 7 · 3⤊ 1⤋
a since these values will make the bottom zero. Any number over 0 is infinity, which is why a vertical asymptote approaches infinity.
Then the second part...
Assuming the whole first part is in ( ), find what makes 3x^2+2x-1=0, which is -1 and 1/3, but 1/3 isnt an answer. So a.
2007-12-16 18:42:47 · answer #2 · answered by Illiniguy 2 · 0⤊ 1⤋
vertical asymptotes: set the denominator = 0 by factoring.
a. x = 2 and x = -2
horizontal asymptotes: when the degree of the top is equal to the degree of the bottom, just take the ratio of leading coefficients.
b. y = 3
2007-12-16 18:40:37 · answer #3 · answered by mathgoddess83209 3 · 0⤊ 1⤋
vertical asymptotes when you divide by zero.
Horizontal asymptote is y = lim{x-->infinity} g(x)
2007-12-16 18:40:16 · answer #4 · answered by a²+b²=c² 4 · 0⤊ 0⤋