Posted by Danielle on Thursday, February 26, 2009 at 10:42pm.
You can reduce the number of unkowns to 2 (a and b) by rewriting the equatiosn with substitutions for c and d.
The system appears overdetermined. You could write four equations in two unknowns knowing the value of two points and the fact that f'(x) = 0 at those points. I would try to use the facts that
f'(x) = 3ax^2 + 2bx -12b = 0 at x = -3 and x=2 to solve for a and b.
27a -6b -12b = 27a -18b = 0
12a +4b -12b = 12a -8b = 0
Those last two equations are equivalent; you can only use one of them. So you need to use an equation that says f(-2) = 0, for example.
8a + 4b -[12b*(-2)] +3 -27b = 0
8a + b = -3
24a + 3b = -9
24a -16b = 0
19 b = -9
This does not lead to an integer value for b. I may have made a mistake somewhere. Check my work and thinking
I understand everything up until the last part with the using equations. Where did 8a and 4b come from?
I too got myself all caught up in nasty fractions, so I tried a different approach.
Since (2,0) is a minimum and it touches the x-axis, there has to be a double root at x=2
so we have (x-2)(x-2) as factors
since it is a cubic, there can be only one other linear factor
so let the function be
y = a(x-b)(x-2)^2 , my b is not necessarily the same as the drwls's b)
I then found the derivative of that and subbed in the fact that if x=-3, dy/dx = 0
this led to 55a + 10ab = 0
a(55 + 10b) = 0
a = 0, not possible if we want a cubic
or
b = -11/2
I then subbed in (-3,3) in y = a(x-b)(x-2)^2 and using the fact that b=-11/2 I got
a = 6/125
so my function is
f(x) = (6/125)(x+11/2)(x-2)^2
both of your points (2,0) and (-3,3) satisfy,
I then found the derivative of that function and set it equal to zero
that gave me x = 2, or x = -3
Q.E.D.
I will leave it up to your to expand it.
As drwls also noted, there seems to be redundant information.