For the first question I believe it’s 3,2 and 2,3
Hello,
A cubic polynomial with -2,1,2 as roots is
f(x)=k*(x+2)(x+1)*(x-2)=k(x²-4)(x+1)=k(x^3+x²-4x-4)
Looking the answers ==>k=1
==>Answer A
Put a dot on the 4 on the y line. then from that point go down 3 and to the right 4.
∠A = 180° - (63°+101°)=16°
sine's rule
<span>First of all, F is true because (16 - 16)/(16 - 4) = 0. It just means when you plug in that number (16) you'll get 0. This will NOT happen for 4 because the 0 is in the denominator.
Look what happens if you have x = 4. This gives you 12/0, which is undefined. For some graphs, this is a hole, but let's look closer.
What happens if you have x = 3.9? You'll have 12.1/0.1. 3.9999? 12.0001/0.0001. The closer you get to 4, the closer you'll get to y = infinity.
But what if you have 4.1? 11.9/-0.1. You'll get the same results, but NEGATIVE infinity. So it is NOT a hole in the graph.
If you draw it out, you'll see that there is a vertical asymptote at x = 4.
B and F are true.
As for horizontal asymptotes, look at it like this: y = 16-16/16 - 4 means y = 0. There is no asymptote here. Try subbing in 1 = (x -16)/(x - 4).
Multiply by x - 4 on both sides
x - 4 = x - 16
There is no solution here; there will be an asymptote, so D is also true.
B, D, and F are true. ,"yahoo answers"//////////////////if i get busted at least i put down my main source (you know what I'm saying)</span>
