1. The problem statement, all variables and given/known data Given that (visually) there is a polynomial with a root at x=5 and seems like a triple root at x=1. It cuts the y-axis at y=-10 and there is a max turning point at (3,32). Ok so basically, find the equation of the function P(x) given: P(0)=-10 P(5)=0 P(1)=0 (triple root, thus cubic shape) P'(3)=0 3. The attempt at a solution Judging by the roots of the polynomial, I'm assuming the equation will be of the form: [tex]P(x)=a(x-1)^3(x-5)[/tex] When I substitute (0,-10) I get a=-2 Thus, [tex]P(x)=-2(x-1)^3(x-5)[/tex] I tested the point (3,32) and since it satisfied the equation when I did it in the test, I assumed it was the turning point; but when I graphed it at home, I found that the turning point wasn't at (3,32) but just seemed to pass through that point. When I take the derivative of [tex]P(x)=a(x-1)^3(x-5)[/tex] I get [tex]P'(x)=4a(x-1)^2(x-4)[/tex] and so P'(3)=0 only when a=0, but that is obviously not the answer I want for a. Maybe for what I assumed the form of the polynomial P(x) is incorrect? Is it possible to find an equation to satisfy all these criteria? Honestly, being that this question was only worth 2 marks out of a 75 mark exam that was 90 mins long, I believe it could've just been a parabola that was hand-drawn really badly... Anyway, thanks for any help or advice on how to approach this.