Question 7705
you have 3 roots, at x=1, x=3 and x=6, so we are taking about a cubic, where (x-1)(x-3)(x-6) = 0. The following graph shows just 3 of the possible curves:


{{{graph(200, 200, -1, 7, -10, 25, (x-1)(x-3)(x-6), 5(x-1)(x-3)(x-6), 0.4(x-1)(x-3)(x-6))}}}


the issue is now to find that ONE curve that passes through (4, -12). All the possible curves are just "multiples" of y = (x-1)(x-3)(x-6).


So, we have y = a(x-1)(x-3)(x-6), where a is a constant, so now put in the x and y values.


-12 = a(4-1)(4-3)(4-6)
-12 = a(3)(1)(-2)
-12 = -6a
--> a = 2


so, our required curve is y = 2(x-1)(x-3)(x-6)


jon.