I’m studying high school math and encountered this question in the extension section for derivatives. The text says an algebraic solution to the problem is harder but possible. There is also a similar question:
Show that every point on the plane of the graph $y = ax^3 + bx^2 + cx + d$ lies on at least one tangent to this cubic.
I would appreciate any solutions/methods that are within the grasp of high school math.
On
Here is an approach that you can take to find a solution.
Find, in terms of a variable $a$, the slope of the tangent line to the curve at the point $(a,a^3)$.
Find the slope-intercept equation of the tangent line at $(a,a^3)$ in terms of $a$.
Repeat this for a diferent point $(b,b^3)$
Find, in terms of $a$ and $b$ the coordinates of the two tangent lines. Simplify the coordinates as much as possible. For example, recall how to factor $a^2-b^2$ and $a^3-b^3$.
Suppose that there is a third point $(c,c^3)$ whose tangent line intersects the tangent line for $(b,b^3)$ at the same intersection point found for $a$ and $b$.
Equate the $y$-coordinate of the $b,c$ intersection with the $y$-coordinate of the $a,b$ intersection and solve the equation for $c$ in terms of $a$ and $b$.
Pick two random values for $a$ and $b$ and compute the corresponding value of $c$. With a bit of trial and error you should be able to find integer values of $a$ and $b$ which yield an integer value for $c$.
Use the formulas to find the intersection of tangent lines corresponding to $a$ and $b$, $a$ and $c$, $b$ and $c$ to verify that they all intersect in the same point.
Note: There might be a simpler way, but this in one which is within the skill set of a high school student.
On
$\text{Here is another approach}\\ P(t,t^3)\\ \text{Tangent to} \ y=x^3 \ \text{ at P......(using point-slope form)} $ $$(y-t^3)=3t^2(x-t)$$ $\text{Let's say this passes through} \ H(X_0,Y_0) \\ \text{Cubic simplifies to}$ $$t^3-\frac{3X_0}{2}t^2+\frac{Y_0}{2}=0$$ $\text{Now for it to have 3 real solutions the product of ordinates of the turning points of this equation must be negative }$ $$f'(t)=3t^2-3X_0t$$ $\text{Turning points-}$ $$S\left(0,Y_0\right) \ and \ R\left(X_0,\frac{Y_0-{X_0}^3}{2}\right) $$ $\text{Hence region will be}$ $$Y_0\left(Y_0-X_0^3\right) \lt 0$$
If (as in the other answer) you find two different tangents for parameter values $a\neq b$ of the parametrization $(t,t^3),$ they intersect in $$(\frac23(a+b-\frac{ab}{a+b}),-2 a b (-\frac{ab}{a+b}))\quad (\star),$$ then you also find a third different tangent for $c=-\frac{ab}{a+b}$: $(\frac23 (a+b+c),- 2 a b c).$ You need the second question answered (see below) to conclude all points of the plane are on a tangent for $y-x^3=0,$ so the ones not of the form $(\star)$ have exactly one tangent to the curve through them. And $(\star)$ or $(\frac23(\frac{a^2+ab+b^2}{a+b}),2\frac{a^2b^2}{a+b})$ only fills the first quadrant below the curve and the third quadrant above the curve, as we can see by $$(\zeta-a)(\zeta-b)(\zeta-c)=\zeta^3-(a+b+c) \zeta^2+ (ab+ac+bc)\zeta-abc\\=\zeta^3-\frac32 x_0 \zeta^2+0\zeta+\frac12 y_0$$ having discriminant $$-27 y_0 (y_0-x_0^3),$$ and remembering that the discriminant is zero if and only if at least two roots are equal. If the coefficients are real numbers, and the discriminant is not zero, the discriminant is positive if the roots are three distinct real numbers, and negative if there is one real root and two complex conjugate roots (see wikipedia).
For the second question: For $y = ax^3 + bx^2 + cx + d$ the tangent at $(t,at^3 + bt^2 + ct + d)$ is $y-(3at^2+2bt+c)x+2at^3+bt^2-d=0$ i.e. for a given $(x_0,y_0)$ we get $x = x_0, 0 = -2at^3+(3x_0a-b)t^2+2x_0bt+d+x_0c-y_0=p(t)$ we only need the existence of a root and since this is a degree three polynomial in $t$ we always find a parameter value for which there is a tangent though $(x_0,y_0)$ by the MVT and the fact that for sufficiently big values of $R,$ $p$ has different sign of the values $p(-R)$ and $p(R)$.