I need to find b, c and d in a cubic equation of the form y=x3+bx2+cx+d (ie. a=1 if a is the coefficient of x3). I've been given three points (-2,-13), (-1,0), (1,2). If I had a fourth piece of data i could solve the equation I suppose, but with only three I've really struggled. Here's what I've got so far: (1) is just the set out (2) I've entered the data (3) I've rearranged the equation But when I thin input the equation on my calculator (the TI-89 Cas calc) I get a matrix in which every row has x and y in it. But that solution doesn't agree with the back of the book, which states that: b=-2 c=0 d=3 which indicates a final equation of y=x3-2x2+3 I've thought of using a matrix system which ommits the top row of all three matrices, but of course that equation system isn't defined. I'd really apprieciate any help, thanks.