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Struggling with this one:

Find the equation with the general form:
$$f(x)\to ax^3+bx^2+cx+d=0$$ $$f(x)'\to 3ax+2bx+c=0$$ $$f(x)''\to 6ax+2b=0$$

Points given:
- curve cuts x-axis at (-3|0)
- curve has high/low point at (-2|?)
- curve has infliction point at (-2/3|?)

Solution is given just for self correction: $$f(x)\to x³+2x²-4x-3=0$$

I defined three equations but fail to find the fourth one: $$(I)\to f(x)\to -27a+9b-3c+d=0$$ $$(II)\to f(x)'\to 12a-4b+c=0$$ $$(III)\to f(x)''\to -4a+2b=0$$

Is my approach wrong to find the fourth equation so I can calculate the four unknowns?

Edit:
The suggestion is that the fourth equation should be 1: $$(IV)\to f(x)'''\to 6a=0$$ $$(IV)\to f(x)'''\to 6a=6$$ $$(IV)\to f(x)'''\to a=1$$

This gives the following values for the solver:
(I) -27 | 9 | -3 | 1 | 0
(II) 12 | -4 | 1 | 0 | 0
(III) -4 | 2 | 0 | 0 | 0
(IV) 1 | 0 | 0 | 0 | 1

Result: a=1 b=2 c=-4 d=-3

tobias47n9e
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  • express $a,b,c$ in terms of $d$ and then put their values in $f(x)$ then cancel out $d$ assuming $d\ne0$ – lab bhattacharjee Jan 11 '14 at 11:19
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    For an infliction point, you'll want $f'''(-\frac23) = 6a \neq 0$ Therefor $a$ can be divided out and $a=1$ can be chosen. – AlexR Jan 11 '14 at 11:29
  • @lab-bhattacharjee: I'm not sure which equation you mean. Where does d cancel out? – tobias47n9e Jan 11 '14 at 11:59
  • @AlexR: Also not sure where a would divide out. I'm really so confused now by this, that I'm not sure which equation to manipulate. – tobias47n9e Jan 11 '14 at 12:02
  • @Spießbürger, I didn't use "equation". – lab bhattacharjee Jan 11 '14 at 12:07
  • @labbhattacharjee: My first thought was -a=(-9b+3c-d)/27 But as I said, I'm confused. Hope you understand. – tobias47n9e Jan 11 '14 at 12:12
  • From the formulation of the question and the given solution I would assume that $a=1$ is given as fixed. The conditions on the cubic do not depend on the scale. – Lutz Lehmann Jan 11 '14 at 12:27
  • @LutzL: The problem does not work on the assumption that a is 1. I am not sure how to say that in English, but maybe a tangent or linear factor has to substituted for one of the values – tobias47n9e Jan 11 '14 at 12:30
  • @Spießbürger What LutzL as well as me tried to point out is that you fourth equation can be arbitrarily chosen $a=\text{const}$ with any nonzero constant. Thus we recommend $a=1$ as the fourth equation and the system should then become uniquely solvable. – AlexR Jan 11 '14 at 13:27

1 Answers1

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We have already established $a=1$ from the comments.

The three equations should be:

  • $-27a+9b-3c+d=0$
  • $12a-4b+c=0$
  • $-4a + 2b = 0$

Do you see where you made a slight algebra error in the third equation?

Now, solving these in reverse order with $a=1$, yields:

  • $b = 2$
  • $c = -4$
  • $d = -3$.

Of course, now you should find the missing values for the points.

Amzoti
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  • Still having a little trouble comprehending the a=1 step. Thank you anyway for your detailed answer. I will accept this and revisit this after reviewing some other polynom chapters. – tobias47n9e Jan 12 '14 at 19:47
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    From the third equation, you have $b = 2a$, so you can try $a = 1, a = 2$ and see if it changes the result. Regards – Amzoti Jan 12 '14 at 19:52