When given a cubic graph with one real root. I need to find the equation of that graph using the function $$y=a(x-s)(x^2+bx+c)$$ where a, b,and c are unknowns. The y intercept is therefore $t = -asc$ Cubic with indicated points $$\frac{dy}{dx} = a(3x^2+2bx-2s x - sb +c)$$
There are 2 equations and 3 unknowns. I know that I could draw a tangent from point $s$ to arbitrary point R on the graph. This allows me to say that that line has an equation $$y= m(x-s)$$ Therefore I can say: $$a(x-s)(x^2+bx+c)= m(x-s)$$ at point R.$$m = a(x^2+bx+c)$$ Then ( and this is the part I don't quite understand) "at the tangent there are equal roots, so that means that $b^2-4ac = 0$"
How do I continue solving this problem? Does it have a clear solution?