8. Here is the problem…..Determine the rational function R(x) has a graph that crosses the x-axis at -1, touches the x-axis at -4, has vertical asymptotes at x=-2 and x=3, and has one horizontal asymptote at y=-2……
Is this the answer? R(x)=(-2(x+1)(x+4))/(x+2)(x-3)
2007-07-16 11:21:03 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You are almost right, except x+4 should
be (x+4)². If the graph touches the x-axis at -4,
it must have a multiple zero there.
2007-07-16 11:33:27 · answer #1 · answered by steiner1745 7 · 1⤊ 0⤋
Let's just check if your answer is correct :
The asymptotes at -2 and 3 are there.
R(-1) = 0, OK.
R(-4) = 0, OK.
You also need an horizontal tangent at -4 ("touches the horizontal axis", without crossing it).
R'(x) = (x+2)(x+3)(-4x+5) + 2(x+1)(x+4)(2x-1) / (x+2)^2*(x-3)^2
R'(-4) = -22 / ...
Nope, it doesn't work...
I'll try to come up with another solution for you. Please hold on.
2007-07-16 18:33:43 · answer #2 · answered by Kilohn 3 · 0⤊ 0⤋
Man, sorry I wish I could help on this one. Maybe the site I listed below? Good luck!
2007-07-16 18:24:31 · answer #3 · answered by FridaY 3 · 0⤊ 0⤋