I am trying to find the equation of a quartic function given its three turning points. Here is the information I have: maximum turning points at $(-12, 1)$ and $(-6,1)$; minimum turning point at $(-9, -1)$; points at $(-15,-1)$ and $(-3,-1)$.
I have tried substituting the $5$ points into the equation $y=ax^4+bx^3+cx^2+dx+e$, but that didn't work. I also read somewhere that the derivative of a quartic function is the cubic equation which I think means at the turning points it is equal to zero as the gradient of the tangent is zero.
I managed to solve a similar problem with two turning points of a cubic function using $f'(x)=3ax^2+2bx+c$ and then simultaneous equations but using the derivative and simultaneous equations didn't work for the quartic.
From what I've read it might have something to do with calculus but I'm hoping there's a simpler solution as I don't even know what calculus is.
I only have beginner level knowledge of polynomials and functions so any help would be very much appreciated :).
Thanks very much,
Harry
pretty clear the derivative is a multiple of $(x+12)(x+9)(x+6),$ so $$ f' = A \left(x^3 + 27x^2 + 234x + 648 \right) $$ $$ f' = A x^3 + 27 A x^2 + 234 A x + 648 A $$ $$ f = \frac{A}{4} x^4 + 9 A x^3 + 117 A x^2 + 648 A x + C $$