Additional Mathematics 4037/23 — October/November 2013

Walkthrough

A stationary point is where the gradient of the curve is zero, so we begin by differentiating. Using the power rule term by term on $y = x^3 - 6x^2 - 36x + 16$ gives $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 12x - 36$ (the constant differentiates to 0).

Setting this equal to zero gives a quadratic in $x$. Dividing by 3 simplifies it to $x^2 - 4x - 12 = 0$, which factorises as $(x-6)(x+2) = 0$, so $x = 6$ or $x = -2$. These are the $x$-coordinates of the two stationary points.

To find the corresponding $y$-coordinates we substitute each value back into the original curve equation (not the derivative): at $x = 6$, $y = 216 - 216 - 216 + 16 = -200$; at $x = -2$, $y = -8 - 24 + 72 + 16 = 56$.

Key Takeaways

• Stationary points occur where $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.
• Differentiate term by term using the power rule; constants vanish.
• Always substitute back into the original equation to get the full coordinates.

Common Mistakes

• Dropping a sign when differentiating: $-36x$ differentiates to $-36$, not $+36$; the mark scheme awards B2 with only B1 if two terms are correct.
• Substituting into the derivative instead of the original equation when finding the $y$-values.
• Sign slips when evaluating: e.g. computing $y(-2)$ as $-8 - 24 - 72 + 16$ instead of $-8 - 24 + 72 + 16$.
• Forgetting that both roots must be found — the question asks for all stationary points.

Things to Be Careful About

• The mark scheme requires the correct three-term derivative for full credit; check every coefficient before moving on.
• Both coordinate pairs must be given — one pair alone earns only part of the accuracy marks.
• Answers should be given as exact integer coordinates here; no rounding is involved.