Find equation of cubic from turning points

Question

If I have a cubic where I know the turning points, can I find what its equation is?

I already know that the derivative is 0 at the turning points. Other than that, I'm not too sure how I can continue.

Suppose I have the turning points (-2,5) and (4,0). I have started doing the following: $$ \frac{dy}{dx} = 0 \text{ at turning points}\\ \text{So, } 0 = (x+2)(x-4)\\ $$ $$ \int (x+2)(x-4) dx\\ = \int x^2-2x-8 dx\\ = \frac{x^3}{3} - x^2 - 8x + C $$

Graphing this, you get correct $x$ coordinates at the turning points, but not correct $y$.

I'm aware that only with that information you can't tell how steep the cubic will be, but you should at least be able to find some sort of equation.

Finally, would a $y$-intercept be helpful? If so, then suppose for the above example that the $y$-intercept is 4.

Answer 1

Welcome! You simply forgot that having the turning points provides the derivative up to a nonzero constant factor, i.e. $$y'(x)=K(x+2)(x-4),\quad K\in \mathbf R^*, \quad \text{ whence }\;y(x)=K\biggl(\frac{x^3}3 -x^2-8x\biggr)+C.$$ You now have two constants to adjust the ordinates.

Answer 2

If the turning points of a cubic polynomial $f(x)$ are $(a, b)$ and $(c, d)$ then $f(x) =k(\dfrac{x^3}{3}-\dfrac{(a+c)x^2}{2}+acx)+h $ where $k =-\dfrac{6(b-d)}{(a-c)^3} $ and $h =\dfrac{b+d}{2}-\dfrac{(b-d)(a+c)(a^2+c^2-4ac)}{2(a-c)^3} $.

Here's the details.

If the turning points are $(a, b)$ and $(c, d)$ then $f'(x) =k(x-a)(x-c) =k(x^2-(a+c)x+ac) $.

Then $f(x) =k(\dfrac{x^3}{3}-\dfrac{(a+c)x^2}{2}+acx)+h $.

$f(a) =k(\dfrac{a^3}{3}-\dfrac{(a+c)a^2}{2}+a^2c)+h =b $ and $f(c) =k(\dfrac{c^3}{3}-\dfrac{(a+c)c^2}{2}+ac^2)+h =d $.

Subtracting,

$\begin{array}\\ b-d &=k(\dfrac{a^3-c^3}{3}-\dfrac{(a+c)(a^2-c^2)}{2}+ac(a-c))\\ &=k(a-c)(\dfrac{a^2+ac+c^2}{3}-\dfrac{(a+c)(a+c)}{2}+ac)\\ &=k(a-c)(\dfrac{a^2+ac+c^2}{3}-\dfrac{(a+c)(a+c)}{2}+ac)\\ &=k(a-c)(\dfrac{a^2+ac+c^2}{3}-\dfrac{a^2+2ac+c^2}{2}+ac)\\ &=k(a-c)(\dfrac{2(a^2+ac+c^2)-3(a^2+2ac+c^2)+6ac}{6})\\ &=k(a-c)(\dfrac{-a^2-c^2+2ac}{6})\\ &=-k(a-c)(\dfrac{(a-c)^2}{6})\\ &=-k\dfrac{(a-c)^3}{6}\\ &=k\dfrac{(c-a)^3}{6}\\ \end{array} $

so $k =-\dfrac{6(b-d)}{(a-c)^3} $.

To maintain symmetry, I'll add the two equations.

$\begin{array}\\ b+d &=k(\dfrac{a^3+c^3}{3}-\dfrac{(a+c)(a^2+c^2)}{2}+ac(a+c))+2h\\ &=k\dfrac{2(a^3+c^3)-3(a+c)(a^2+c^2)+6ac(a+c)}{6}+2h\\ &=k\dfrac{2(a^3+c^3)-3(a^3+ac^2+a^2c+c^3)+6a^2c+6ac^2}{6}+2h\\ &=k\dfrac{-a^3-c^3+3a^2c+3ac^2}{6}+2h\\ &=k\dfrac{-(a^3+c^3)+3ac(a+c)}{6}+2h\\ &=k\dfrac{-(a+c)(a^2-ac+c^2)+3ac(a+c)}{6}+2h\\ &=k\dfrac{-(a+c)(a^2+c^2-4ac)}{6}+2h\\ &=-\dfrac{6(b-d)}{(a-c)^3}\dfrac{-(a+c)(a^2+c^2-4ac)}{6}+2h\\ &=\dfrac{(b-d)(a+c)(a^2+c^2-4ac)}{(a-c)^3}+2h\\ \end{array} $

so $2h =(b+d)-\dfrac{(b-d)(a+c)(a^2+c^2-4ac)}{(a-c)^3} $ and $k =-\dfrac{6(b-d)}{(a-c)^3} $.

Answer 3

If given points $(x1,y1)$ and $(x2,y2)$,

we can use the fact that the roots of the derivative are $x1$ and $y1$(simplifying everything else for now)

$f'(x)=(x-x1)(x-x2)$

and multiply out:

$f'(x)=x^2-(x2+x1)x+x2x1$

take the integral

$f(x)=∫f'(x)=x^3/3-(x2+x1)x^2/2+x2x1x+k$

simplify to a case where $x1$ and $y1 = 0$

$f(x)=x^3/3-x2x^2/2$

multiply function by a correction factor

$a=y2/f(x2)$

$g(x)=a*f(x)$

to add other values for $x1$ and $y1$, $x2$ to $x2-x1$, $y2$ to $y2-y1$

$h(x)=x^3/3-(x2-x1)x^2/2$

$b=(y2-y1)/h(x2-x1)$

change $x$ to $x-x1$, add $y1$ value to fit the curve:

$i(x)=b*h(x)+y1$

simplify

find $b$

$h(x2-x1)=(x2-x1)^3/3-(x2-x1)(x2-x1)^2/2$

$h(x2-x1)=(x2-x1)^3/3-(x2-x1)^3/2$

$h(x2-x1)=2(x2-x1)^3/6-3(x2-x1)^3/6$

$h(x2-x1)=-(x2-x1)^3/6$

$b=(y2-y1)/(-(x2-x1)^3/6)$

$b=-6(y2-y1)/(x2-x1)^3$

multiply $h(x-x1)$ by $b$

$i(x)=(-6(y2-y1)/(x2-x1)^3)((x-x1)^3/3)-(-6(y2-y1)/(x2-x1)^3)((x2-x1)x^2/2)$

Simplify

$i(x)=-2(y2-y1)(x-x1)^3/(x2-x1)^3+3(y2-y1)(x-x1)^2/(x2-x1)^2$

Add $y1$

$i(x)=-2(y2-y1)(x-x1)^3/(x2-x1)^3+3(y2-y1)(x-x1)^2/(x2-x1)^2+y1$

it isn't possible to do this with a quartic as there are 5 coefficients to solve for(in standard form) and getting equations with $(x1,y1)$, $(x2,y2)$, $(x3,y3)$ with each point satisfying the equation and each x-value being a root of the derivative, there are 6 distinct equations in a,b,c,d,and e, meaning there is not solution unless there is a special case.