Tangent and Normal to a Cubic Curve

This is a standard Paper 1 opener: it rewards clean differentiation and knowing that a normal is perpendicular to the tangent. Every mark below is a method or accuracy mark you can secure without a calculator.

The curve $y = 2x^3 - 5x^2 + 1$. (i) Find $\dfrac{dy}{dx}$. [2] (ii) Find the equation of the tangent to the curve at the point where $x = 2$. [4] (iii) Find the equation of the normal at the same point, giving your answer in the form $ax + by + c = 0$. [3]

The working

(i) Differentiate term by term, multiply by the power then drop it by one: $\frac{dy}{dx} = 6x^2 - 10x \quad \text{(M1 for one correct term, A1 all correct)}$

(ii) The tangent needs a gradient and a point.

Gradient at $x = 2$: substitute into $\frac{dy}{dx}$: $\frac{dy}{dx} = 6(2)^2 - 10(2) = 24 - 20 = 4 \quad \text{(M1)}$

The point: substitute $x = 2$ into the original curve (not the derivative): $y = 2(2)^3 - 5(2)^2 + 1 = 16 - 20 + 1 = -3 \quad \text{so the point is } (2, -3) \text{ (A1)}$

Tangent line through $(2, -3)$ with gradient $4$: $y - (-3) = 4(x - 2) \;\Rightarrow\; y = 4x - 11 \quad \text{(M1, A1)}$

(iii) A normal is perpendicular to the tangent, so its gradient is the negative reciprocal: $m_{\text{normal}} = -\frac{1}{4} \quad \text{(M1)}$

Through the same point $(2, -3)$: $y + 3 = -\frac{1}{4}(x - 2)$

Multiply through by $4$ and collect into the requested form: $4y + 12 = -(x - 2) \;\Rightarrow\; x + 4y + 10 = 0 \quad \text{(M1, A1)}$

Where the marks are won and lost

• The y-coordinate in (ii) is a separate accuracy mark. Students who use $(2, 0)$ or skip the point lose it every time.
• The instruction “in the form $ax + by + c = 0$” is a mark. Leaving the normal as $y = -\frac14 x - \frac52$ when a specific form was asked can cost the final A1. Read the demanded form and match it exactly.
• Negative-reciprocal, not negative. The normal gradient of a line with gradient $4$ is $-\frac14$, not $-4$.

Common mistakes

• Substituting $x = 2$ into $\frac{dy}{dx}$ to get the point’s $y$-value (that gives the gradient, not $y$).
• Using $+\frac14$ or $-4$ for the normal gradient.
• Forgetting to clear fractions when a form like $ax + by + c = 0$ with integer coefficients is expected.

The routine in full: Tangents & Normals notes. For the wider topic, see the Calculus pillar.