f(x)=x^4-5x^3-50x^2/6x+2
given that find the vertical asymptote and horizontal asymptotes
find the x-intercepts
PLEASE find the following and you will get 10 points
2007-10-05 10:44:23 · 3 answers · asked by star baller 360 5 in Science & Mathematics ➔ Mathematics
the verital asymptotes is wrong it not -3 please find the vertical asymptotes
2007-10-05 10:58:18 · update #1
y = (x^4 - 5x^3 - 50x^2) / (6x + 2)
Is this what you meant to write?
y = x^2 * (x-10)*(x+5) / 2*(3x + 1)
There is a vertical assymptote at x = -1/3
To find the horizontal assymptotes, find limit as x --> infinity
y --> x^4 / 6x = x^3 / 6
y --> inf as x --> inf
y --> -inf as x --> -inf
There is no horizontal assymptote.
2007-10-05 11:21:54 · answer #1 · answered by Dr D 7 · 0⤊ 0⤋
Jenh4200 assumes you have 50x^2/(6x+2)
and set (6x+2) = 0
if your eq is 50x^2/6x then x=0 is the vertical asymptote
Vertical asymptote is found when there is a fractional term and you set the demo to 0
2007-10-05 18:19:00 · answer #2 · answered by norman 7 · 0⤊ 0⤋
Vert As: Set 6x + 2 = 0, so x = -3
Oops, no, should be x = -1/3
Horiz: None, since higher power is on top
x-intercepts: Set top = 0
Factor out x^2
x^2(x^2 - 5x - 50) = 0
x^2 (x-10)(x+5) = 0
x = 0, 10, and -5
These are the x-intercepts.
2007-10-05 17:48:36 · answer #3 · answered by jenh42002 7 · 0⤊ 0⤋