Determine the equation of a cubic function that goes through the point (1,7) and has a zero at x=-1 and another zero at x=3 that is also a turning point.

Answer 1

y k = (x-a)(x-b)(x-c)
zero at x = -1
so let a = -1
y k = (x+1)(x-b) (x-c)
zero at = 3
so let b = 3
y k= (x+1) (x-3)(x-c)
if x = 1, y = 7
7 k = 2 (-2)(1-c)
-7k / 4 = 1-c
c = 1 + 7k / 4
y k= (x+1) (x-3)(x-1 -7k/4)
y = (x^2 -2 x - 3)(x-1-7k/4)
at x = 3, dy/dx = 0
dy/dx = (x^2 -2 x - 3)*1 + (x-1-7k/4)(2x-2)
0 = (9-6-3) + (2-7k/4)(4) = 0 + (2-7k/4)(4)
so 7 k/4 = 2
k=8/7
c = 1 +7 k/4 = 1 + 7(2/7) = 3
y k = (x+1)(x-3)(x-3) (of course you get two y values at the turning point, just thought of that, the drivative stuff was unnecessary)
8/7 y= (x+1)(x-3)(x-3) you multiply it out

Answer 2

y=(x-3)(x+1)(x-(11/4))