F(x)=(x^2+5x-24)/(3x^2-5x-28)
So far, all I've gotten is for (x^2+5x-24) to equal (x-3)(x+8) and now I'm stuck. Any help is appreciated. Thanks.
2007-09-21 10:28:52 · 3 answers · asked by Glenn M 1 in Science & Mathematics ➔ Mathematics
To get horizontal asymptotes, you need the limit as x goes to +/-infinity. This 1/3, so the asymptote is y = 1/3.
To get the veritcal asymptotes, factor the denominator to get:
(3x +7)(x -4). It is 0 at x=4,-7/3 so these are your vertical asymptotes.
Factoring the numerator gets you the zeros of the function (where it crosses the x-axis).
2007-09-21 10:38:15 · answer #1 · answered by Phineas Bogg 6 · 0⤊ 0⤋
f(x) = [(x-3)(x+8)]/[(3x+7)(x-4)] There will be vertical asymptotes at x = 4 and x = -7/3 since thos values make the denominator = 0.
There is a horizontal asymptote at y =1/3 since x^2/3x^2 = 1/3.
2007-09-21 10:51:31 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
HA (horizontal asymptote) occurs where f (x) = 0, namely where the numerator of the derivative = 0
VA occurs where f (x) is undefined, namely where the denominator of the derivative =0
denominator (3x^2 +5x -28) = (3x + 7)(x - 4)
2007-09-21 12:13:58 · answer #3 · answered by intrepid_mesmer 3 · 0⤊ 0⤋