Anyone? Can you please just check the second one for me?
Determine an equation in factored form for the polynomial represented by the graph.
Graph #1: Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
Looking at this I can say:
>4 degree polynomial; Quartic
>Zero's: -1, +1, 2
>Even Degree Polynomial
>Point at: (0,4)
y= a (x+1) (x-1) (x-2)
...plug in (0,4) -> 4=a(0+1)(0-1)(0-2) -> 2=a
y=2(x+1)(x-1)(x-2)................Is this answer correct? Shouldn't this have been an Even Degree Polynomial?(Also just making sure, is this in factored form?)
Graph #2. Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
Looking at this I can say:
>4 Degree Polynomial (minimum); Quartic
>Zero's: -2, +1
>Point at: (0,-2)
y=a (x+2) (x-1)
...plug in (0,-2) -> -2=a (0+2)(0-1) -> 0=a -> ?
y=(x+2)(x-1)......................I'm certain this is unfinished/incorrect
.........................................Since a=0, and no other point in defined on the graph...? I don't know how I would solve this?
First problem: there must be four roots. since you have only three roots, the fourth cannot be complex because complex roots are in pairs of congugates. The fourth root must be pont 2 as a double root. To get the correct answer you have to square the binomial for the double root, i.e. (x-2)^2
do the same in problem 2