Recover a cubic from its extrema · source reference

Printed question

The cubic curve $C$ has a maximum point $A$ and a minimum point $B$. The line segment $AB$ is the diameter of a circle with equation $(x-5)^2+y^2=25$. Given that the $x$-coordinate of $B$ is 1, find the equation of $C$.

[2011/TJC/JCT/JC1/4]

Printed working

Since $AB$ is the diameter of the circle with centre $(5,0)$ and the $x$-coordinate of $B$ is 1, then the $x$-coordinate of $A$ must be 9.

When $x=1$, $(1-5)^2+y^2=25\Rightarrow y=-3$ or $3$.

Reject $y=3$ because if $B$ is the point $(1,3)$, then $A$ will be lower than $B$ which is not possible since $B$ is the minimum point of the cubic curve.

Therefore, $B$ has coordinates $(1,-3)$ and $A$ has coordinates $(9,3)$.

Let $y=ax^3+bx^2+cx+d$ be the equation of $C$.

$\therefore\frac{dy}{dx}=3ax^2+2bx+c$.

Since max and min pts are $(9,3)$ and $(1,-3)$ respectively, we have

$$\begin{cases}243a+18b+c=0\\729a+81b+9c+d=3\\3a+2b+c=0\\a+b+c+d=-3.\end{cases}$$

Using GC, $a=-0.0234,b=0.352,c=-0.633,d=-2.70$ (to 3 s.f.).

$\therefore y=-0.0234x^3+0.352x^2-0.633x-2.70$.

Teacher notes

Horizontal distance of $A$ and $B$ to centre of circle is 4 units, so $x$-coord of $A=5+4=9$.

1. At max and min pts, $\frac{dy}{dx}=0\Rightarrow\frac{dy}{dx}=0$ at $x=1$ and $9$.

2. Curve $C$ passes through max and min pts $\Rightarrow(9,3),(1,-3)$ satisfy eqn of $C$.

Printed answer key

$y=-0.0234x^3+0.352x^2-0.633x-2.70$.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: EQUATIONS. Question sheet printed pages 15–19; solution sheet printed pages 1–12. Original exam attribution is retained above. Currency dollar signs are spelled out outside maths to avoid confusing currency with TeX delimiters.