IGCSE · Practice questions · Tangents, normals and stationary points

Find it, then prove what it is.

Six original Cambridge-style questions on gradients, tangents, normals and stationary points, with the full justification of maxima and minima that the mark scheme expects.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Differentiate, substitute, then decide.

Point first: if you only have \(x\), find \(y\) from the curve.
Gradient: substitute \(x\) into \(\dfrac{dy}{dx}\); the normal uses \(-\dfrac{1}{m}\).
Line: \(y - y_{1} = m(x - x_{1})\); give the form the question asks for.
Stationary points: solve \(\dfrac{dy}{dx} = 0\) by factorising, then find each \(y\).
Nature: sign of \(\dfrac{d^{2}y}{dx^{2}}\), or the sign of \(\dfrac{dy}{dx}\) either side, with the conclusion written out.

Recall

[3 marks]

Find the equation of the tangent to the curve \(y = 2x^{3} - 5x + 1\) at the point where \(x = -1\).

\(y = -2 + 5 + 1 = 4\), so the point is \((-1, 4)\). ✓
\(\dfrac{dy}{dx} = 6x^{2} - 5 = 6 - 5 = 1\) at \(x = -1\). ✓
\(y - 4 = 1(x + 1)\), so \(y = x + 5\). ✓

Calculation

[4 marks]

Find the equation of the normal to the curve \(y = (2x - 3)^{4}\) at the point where \(x = 1\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.

At \(x = 1\): \(y = (-1)^{4} = 1\). ✓
Chain rule: \(\dfrac{dy}{dx} = 4(2x - 3)^{3} \times 2 = 8(2x - 3)^{3} = 8(-1)^{3} = -8\). ✓
Normal gradient \(= -\dfrac{1}{-8} = \dfrac{1}{8}\): \(y - 1 = \dfrac{1}{8}(x - 1)\). ✓
\(8y - 8 = x - 1\), so \(x - 8y + 7 = 0\). ✓

Calculation

[4 marks]

Find the coordinates of the points on the curve \(y = x^{3} + 3x^{2} - 5x\) at which the tangent is parallel to the line \(y = 4x + 1\).

\(\dfrac{dy}{dx} = 3x^{2} + 6x - 5\). Parallel means gradient 4: \(3x^{2} + 6x - 5 = 4\). ✓
\(3x^{2} + 6x - 9 = 0 \Rightarrow x^{2} + 2x - 3 = 0 \Rightarrow (x + 3)(x - 1) = 0\). ✓
\(x = 1\): \(y = 1 + 3 - 5 = -1\). ✓
\(x = -3\): \(y = -27 + 27 + 15 = 15\). The points are \((1, -1)\) and \((-3, 15)\). ✓

Analysis

[6 marks]

The curve \(y = e^{2x} + 1\) meets the \(y\)-axis at \(P\).

(a) Find the equation of the tangent at \(P\) and the equation of the normal at \(P\).

(b) The tangent and the normal meet the \(x\)-axis at \(A\) and \(B\). Find the area of triangle \(APB\).

(a) At \(x = 0\): \(y = 1 + 1 = 2\), so \(P\) is \((0, 2)\). ✓
\(\dfrac{dy}{dx} = 2e^{2x} = 2\) at \(x = 0\). Tangent: \(y = 2x + 2\). ✓
Normal gradient \(-\dfrac{1}{2}\): \(y = -\dfrac{1}{2}x + 2\). ✓
(b) Tangent at \(y = 0\): \(x = -1\), so \(A(-1, 0)\). Normal at \(y = 0\): \(x = 4\), so \(B(4, 0)\). ✓
Base \(AB = 5\), height \(= 2\) (the \(y\)-coordinate of \(P\)). ✓
Area \(= \dfrac{1}{2} \times 5 \times 2 = 5\) square units. ✓

Proof

[6 marks]

A curve has equation \(y = \dfrac{\ln x}{x}\), for \(x > 0\).

(a) Show that \(\dfrac{dy}{dx} = \dfrac{1 - \ln x}{x^{2}}\).

(b) Find the exact coordinates of the stationary point.

(a) Quotient rule: \[\dfrac{dy}{dx} = \dfrac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^{2}} = \dfrac{1 - \ln x}{x^{2}}\] ✓
(b) \(x^{2} > 0\), so \(\dfrac{dy}{dx} = 0\) when \(1 - \ln x = 0\), that is \(\ln x = 1\), \(x = e\). ✓
\(y = \dfrac{\ln e}{e} = \dfrac{1}{e}\). The stationary point is \(\left(e, \dfrac{1}{e}\right)\). ✓
(c) At \(x = e\): \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{2 - 3}{e^{3}} = -\dfrac{1}{e^{3}}\). ✓
\(-\dfrac{1}{e^{3}} < 0\), so the stationary point is a maximum. ✓
(Check by signs: \(1 - \ln x > 0\) for \(x < e\) and \(< 0\) for \(x > e\), so the gradient goes \(+\) then \(-\).) ✓

Analysis

[7 marks]

The curve \(y = 2x^{3} + ax^{2} + bx\), where \(a\) and \(b\) are constants, has a stationary point at \((1, -5)\).

(a) Find the values of \(a\) and \(b\).

(b) Find the coordinates of the other stationary point.

(a) The point is on the curve: \(2 + a + b = -5\), so \(a + b = -7\). ✓
\(\dfrac{dy}{dx} = 6x^{2} + 2ax + b = 0\) at \(x = 1\): \(6 + 2a + b = 0\), so \(2a + b = -6\). ✓
Subtract: \(a = 1\), then \(b = -8\). ✓
(b) \(\dfrac{dy}{dx} = 6x^{2} + 2x - 8 = 2(3x^{2} + x - 4) = 2(3x + 4)(x - 1)\), so the other root is \(x = -\dfrac{4}{3}\). ✓
\[y = 2\left(-\dfrac{64}{27}\right) + \dfrac{16}{9} + \dfrac{32}{3} = \dfrac{-128 + 48 + 288}{27} = \dfrac{208}{27}\] The point is \(\left(-\dfrac{4}{3}, \dfrac{208}{27}\right)\). ✓
(c) \(\dfrac{d^{2}y}{dx^{2}} = 12x + 2\). At \(x = 1\): \(14 > 0\), so \((1, -5)\) is a minimum. ✓
At \(x = -\dfrac{4}{3}\): \(-16 + 2 = -14 < 0\), so \(\left(-\dfrac{4}{3}, \dfrac{208}{27}\right)\) is a maximum. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.