Home / IGCSE Add Maths 0606 / Calculus / Tangents, normals and stationary points
Cambridge IGCSE · Topic 14.6
IGCSE Add Maths 0606 / Topic 14 / IGCSE

Where the gradient points, and where it stops.

The derivative gives the gradient of a curve at any point. Put in one \(x\)-value and you have the gradient of the tangent there, and so the tangent and the normal. Set the derivative equal to zero and you find the stationary points, where the curve turns; the second derivative, or the sign of the gradient either side, tells you which are maxima and which are minima.

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The key idea

At the point \((x_{1}, y_{1})\) on a curve, the tangent has gradient \(m = \dfrac{dy}{dx}\) evaluated at \(x_{1}\), and equation \(y - y_{1} = m(x - x_{1})\). The normal is perpendicular to the tangent, so its gradient is \(-\dfrac{1}{m}\). Stationary points are where \(\dfrac{dy}{dx} = 0\). If \(\dfrac{d^{2}y}{dx^{2}} < 0\) there, the point is a maximum; if \(\dfrac{d^{2}y}{dx^{2}} > 0\), a minimum. If the second derivative is zero or awkward, check the sign of \(\dfrac{dy}{dx}\) just either side: \(+\) then \(-\) is a maximum, \(-\) then \(+\) is a minimum.

x y O 1 2 3 4 5 1 2 3 4 5 6 7 P(2, 3) maximum (1, 5) minimum (3, 1) y = x³ − 6x² + 9x + 1 tangent at P: y = −3x + 9 gradient −3 normal at P: 3y = x + 7 gradient 1/3
Fig. 1: the curve y = x³ - 6x² + 9x + 1 with its tangent (gradient -3) and normal (gradient 1/3) at P(2, 3). The gradient is zero at the maximum (1, 5) and the minimum (3, 1)
Section 01

Gradients, tangents and normals.

Differentiate, then substitute the \(x\)-coordinate of the point: that number is the gradient of the curve, and of its tangent, at that point. The tangent touches the curve there; the normal passes through the same point at right angles to the tangent (Fig. 1). Two perpendicular gradients multiply to \(-1\), so if the tangent has gradient \(m\) the normal has gradient \(-\dfrac{1}{m}\).

Worked example Tangent and normal at a point
Find the equations of the tangent and the normal to the curve \(y = x^{3} - 6x^{2} + 9x + 1\) at the point \(P\) where \(x = 2\).
  1. \(y = 8 - 24 + 18 + 1 = 3\), so \(P\) is \((2, 3)\).
  2. \(\dfrac{dy}{dx} = 3x^{2} - 12x + 9\). At \(x = 2\): \(12 - 24 + 9 = -3\).
  3. Tangent: \(y - 3 = -3(x - 2)\), so \(y = -3x + 9\).
  4. Normal gradient \(= -\dfrac{1}{-3} = \dfrac{1}{3}\): \(y - 3 = \dfrac{1}{3}(x - 2)\).
  5. Multiply by 3: \(3y - 9 = x - 2\), so \(3y = x + 7\).
Answer: Tangent \(y = -3x + 9\); normal \(3y = x + 7\) (Fig. 1)
Worked example A triangle made by a tangent and a normal
The point \(P(2, 8)\) lies on the curve \(y = x^{2} + \dfrac{8}{x}\). The tangent and the normal at \(P\) meet the \(x\)-axis at \(A\) and \(B\). Find the area of triangle \(APB\).
  1. \(\dfrac{dy}{dx} = 2x - \dfrac{8}{x^{2}}\). At \(x = 2\): \(4 - 2 = 2\).
  2. Tangent: \(y - 8 = 2(x - 2)\), so \(y = 2x + 4\). At \(y = 0\): \(x = -2\), so \(A(-2, 0)\).
  3. Normal gradient \(-\dfrac{1}{2}\): \(y - 8 = -\dfrac{1}{2}(x - 2)\), so \(y = -\dfrac{1}{2}x + 9\). At \(y = 0\): \(x = 18\), so \(B(18, 0)\).
  4. Base \(AB = 18 - (-2) = 20\); the height is the \(y\)-coordinate of \(P\), which is 8.
  5. Area \(= \dfrac{1}{2} \times 20 \times 8\).
Answer: \(80\) square units
Your turn
Find the equation of the normal to \(y = e^{2x}\) at the point where \(x = 0\).
Examiner trap

The normal gradient is \(-\dfrac{1}{m}\): flip AND change the sign. \(\dfrac{1}{m}\) or \(-m\) on its own is wrong. And substitute into \(\dfrac{dy}{dx}\) before writing the line: \(y - 3 = (3x^{2} - 12x + 9)(x - 2)\) is not a straight line.

Section 02

A given gradient, and the stationary points.

Turn the question round: if you know the gradient, set \(\dfrac{dy}{dx}\) equal to it and solve for \(x\). 'Parallel to the line \(y = 2x + 5\)' means gradient 2. The most important case is gradient zero. A point where \(\dfrac{dy}{dx} = 0\) is a stationary point: the tangent there is horizontal, as at the maximum and minimum in Fig. 1.

Worked example Points with a given gradient
Find the coordinates of the points on \(y = x^{3} - 3x^{2} - 7x + 4\) at which the tangent is parallel to the line \(y = 2x - 1\).
  1. Parallel means the same gradient, 2: \(3x^{2} - 6x - 7 = 2\).
  2. \(3x^{2} - 6x - 9 = 0\), so \(x^{2} - 2x - 3 = 0\), so \((x - 3)(x + 1) = 0\).
  3. \(x = 3\): \(y = 27 - 27 - 21 + 4 = -17\).
  4. \(x = -1\): \(y = -1 - 3 + 7 + 4 = 7\).
Answer: \((3, -17)\) and \((-1, 7)\)
Worked example Finding the stationary points
Find the coordinates of the stationary points of \(y = 2x^{3} - 9x^{2} + 12x - 3\).
  1. \(\dfrac{dy}{dx} = 6x^{2} - 18x + 12 = 6(x^{2} - 3x + 2) = 6(x - 1)(x - 2)\).
  2. \(\dfrac{dy}{dx} = 0\) when \(x = 1\) or \(x = 2\).
  3. \(x = 1\): \(y = 2 - 9 + 12 - 3 = 2\).
  4. \(x = 2\): \(y = 16 - 36 + 24 - 3 = 1\).
Answer: \((1, 2)\) and \((2, 1)\)
Your turn
Find the stationary point of \(y = xe^{-x}\).
Examiner trap

Do not cancel \(x\) from \(3x^{2} - 6x = 0\) to get \(x = 2\) only: factorise, \(3x(x - 2) = 0\), and keep \(x = 0\) as well. And finish the job: 'find the stationary points' wants coordinates, not just \(x\)-values.

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Find the gradient of the curve \(y = x^{3} - 4x\) at the point where \(x = 2\).

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The tangent to a curve at a point has gradient \(3\). What is the gradient of the normal at that point?

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At what value of \(x\) does \(y = x^{2} - 6x + 2\) have a stationary point?

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At a stationary point, \(\dfrac{d^{2}y}{dx^{2}} = -5\). What is the stationary point?

Section 03

Maximum or minimum: justify it.

Finding a stationary point is half the question; the other half is deciding its nature, with a reason the examiner can see. There are two valid methods, and you may use either unless the question names one.

MethodMaximumMinimum
Second derivative at the point\(\dfrac{d^{2}y}{dx^{2}} < 0\)\(\dfrac{d^{2}y}{dx^{2}} > 0\)
Sign of \(\dfrac{dy}{dx}\) just left, then just right\(+\) then \(-\)\(-\) then \(+\)

Why it works: at a maximum the gradient falls from positive through zero to negative, so the gradient is decreasing and its derivative, \(\dfrac{d^{2}y}{dx^{2}}\), is negative. At a minimum the gradient increases, so \(\dfrac{d^{2}y}{dx^{2}}\) is positive. If \(\dfrac{d^{2}y}{dx^{2}} = 0\) the test tells you nothing, and you must use the signs of \(\dfrac{dy}{dx}\).

Worked example The second derivative test
Find the stationary points of \(y = x + \dfrac{4}{x}\) and determine their nature.
  1. \(y = x + 4x^{-1}\), so \(\dfrac{dy}{dx} = 1 - 4x^{-2} = 1 - \dfrac{4}{x^{2}}\).
  2. \(1 - \dfrac{4}{x^{2}} = 0 \Rightarrow x^{2} = 4 \Rightarrow x = 2\) or \(x = -2\); then \(y = 4\) or \(y = -4\).
  3. \(\dfrac{d^{2}y}{dx^{2}} = 8x^{-3} = \dfrac{8}{x^{3}}\).
  4. At \(x = 2\): \(\dfrac{d^{2}y}{dx^{2}} = 1 > 0\), so \((2, 4)\) is a minimum.
  5. At \(x = -2\): \(\dfrac{d^{2}y}{dx^{2}} = -1 < 0\), so \((-2, -4)\) is a maximum.
Answer: Minimum \((2, 4)\), maximum \((-2, -4)\). The maximum is LOWER than the minimum: these are local turning points on two separate branches of the curve.
Worked example When the second derivative is zero
Show that \(y = (x - 2)^{4} + 1\) has a stationary point at \((2, 1)\) and determine its nature.
  1. \(\dfrac{dy}{dx} = 4(x - 2)^{3}\), which is 0 at \(x = 2\); then \(y = 0 + 1 = 1\).
  2. \(\dfrac{d^{2}y}{dx^{2}} = 12(x - 2)^{2} = 0\) at \(x = 2\): the second derivative test fails.
  3. At \(x = 1.9\): \(\dfrac{dy}{dx} = 4(-0.1)^{3} = -0.004 < 0\).
  4. At \(x = 2.1\): \(\dfrac{dy}{dx} = 4(0.1)^{3} = 0.004 > 0\).
  5. The gradient changes from negative to positive.
Answer: \((2, 1)\) is a minimum
Your turn
Determine the nature of the stationary point \(\left(1, \dfrac{1}{e}\right)\) of \(y = xe^{-x}\), given that \(\dfrac{dy}{dx} = e^{-x}(1 - x)\).
Examiner trap

Always write the conclusion with its reason, for example '\(\dfrac{d^{2}y}{dx^{2}} = -1 < 0\), so maximum'. A bare value of the second derivative, or 'maximum' with no working, loses the mark. And \(\dfrac{d^{2}y}{dx^{2}} = 0\) does not mean anything on its own: test the sign of the gradient either side.

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

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\(P(4, 4)\) lies on \(y = 2\sqrt{x}\). The tangent and the normal at \(P\) meet the \(y\)-axis at \(A\) and \(B\). Find the area of triangle \(APB\).

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Find the stationary points of \(y = x^{3} - 12x + 5\) and their nature.

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The curve \(y = e^{x} - 2x\) has one stationary point. Find it and state its nature.

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Six original Cambridge-style questions
Tangents and normals at a point, a chain-rule normal, points with a given gradient, a triangle formed by a tangent and a normal, a stationary point of \(\dfrac{\ln x}{x}\), and a cubic with unknown coefficients.