What is 7.3: Tangents, normals and stationary points?

At x=a, the tangent gradient is f'(a) and a non-vertical normal has gradient -1/f'(a); use the point on the curve in point-gradient form. Stationary points solve f'(x)=0.

How many marks are 7.3 questions worth?

7.3 practice drills on this page are worth 3–7 marks each; marks in a real exam vary by question.

Which past exam papers has 7.3 appeared in?

We have not yet indexed a verified past-paper appearance for 7.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

7.3

Apply differentiation to find gradients, tangents and normals, maxima and minima and stationary points, points of inflection; identify where functions are increasing or decreasing.

Draft — not yet indexed

Tangents, normals and stationary points

Worked answers and methods for 7.3 on Edexcel A-level Maths 9MA0.

Explanation

At $x=a$ , the tangent gradient is $f'(a)$ and a non-vertical normal has gradient $-1/f'(a)$ ; use the point on the curve in point-gradient form. Stationary points solve $f'(x)=0$ .
Classify them by a sign change in $f'$ or, when decisive, by $f''(x)>0$ for a minimum and $f''(x)<0$ for a maximum.
A function is increasing where $f'(x)>0$ and decreasing where $f'(x)<0$ .
For a constrained optimisation problem, include endpoints or domain restrictions in the comparison.
A common error is to assume every solution of $f'(x)=0$ is a maximum or minimum; a stationary point can instead be an inflection point.

Worked example

For $f(x)=x^3-6x^2+9x+2$ , find and classify every stationary point. State the intervals on which $f$ is increasing.

1. $f'(x)=3x^2-12x+9=3(x-1)(x-3)$ , so the stationary values are $x=1,3$ .
2.The coordinates are $f(1)=6$ and $f(3)=2$ .
3.Also $f''(x)=6x-12$ , so $f''(1)=-6$ gives a local maximum and $f''(3)=6$ gives a local minimum.
4.The factorised derivative is positive outside the roots, so $f$ is increasing for $x<1$ and $x>3$ .

Answer: Local maximum $(1,6)$ and local minimum $(3,2)$ ; Increasing for $x<1$ and $x>3$

Common mistakes

Don't use $(x,f'(x))$ as the coordinates of a stationary point instead of $(x,f(x))$ .
Don't find stationary x-values and fail to classify them or test the requested increasing intervals.

Exam tip

For stationary-point questions, solve the derivative equation, classify each point and finish with a derivative sign chart.

Worked practice

Tier 1 · Easy

The curve

y=x^2+3x

is considered at the point where

x=1

. Find the equations of the tangent and the normal.

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Tangent $y=5x-1$ Normal $y-4=-\frac15(x-1)$	4
Notes
At $x=1$ , $y=1+3=4$ . Since $\frac{\mathrm dy}{\mathrm dx}=2x+3$ , the tangent gradient is $5$ , giving $y-4=5(x-1)$ and hence $y=5x-1$ . The normal gradient is $-\frac15$ , so $y-4=-\frac15(x-1)$ .

(4 marks)

Tier 2 · Standard

Find equations of the tangent and the normal to

y=x^2+\dfrac{2}{x}

at the point where

x=2

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Tangent: $y-5=\dfrac72(x-2)$ , or any algebraically equivalent equation such as $y=\dfrac72x-2$ Normal: $y-5=-\dfrac27(x-2)$ , or any algebraically equivalent equation such as $y=-\dfrac27x+\dfrac{39}{7}$	5
Notes
At $x=2$ , $y=2^2+2/2=5$ . Differentiate: $dy/dx=2x-2/x^2$ , so the tangent gradient is $4-1/2=7/2$ . Thus the tangent is $y-5=\tfrac72(x-2)$ . The normal gradient is the negative reciprocal, $-2/7$ , so the normal is $y-5=-\tfrac27(x-2)$ .

(5 marks)

Tier 3 · Hard

A model for the volume of an open container is

V=x(12-x)^2

for

0<x<12

, where

V

is measured in

\text{cm}^3

. Use calculus to find the maximum possible volume.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Maximum volume $256\,\text{cm}^3$ , attained when $x=4$	7
Notes
Differentiate the product: $V'=(12-x)^2-2x(12-x)=(12-x)(12-3x)$ . The interior stationary point is $x=4$ ; $x=12$ is outside the open domain as a stationary endpoint factor. Since $V'$ changes from positive to negative at $x=4$ , this is a maximum. Thus $V(4)=4(8)^2=256\,\text{cm}^3$ . Also $V$ tends to $0$ at both ends of the domain, confirming the global maximum.

(7 marks)

Tier 1 · Easy

At the point where

x=2

, determine the tangent to the curve

y=x^3-12x

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$y=-16$	3
Notes
At $x=2$ , $y=8-24=-16$ . Also $\dfrac{\mathrm dy}{\mathrm dx}=3x^2-12$ , which is $0$ at $x=2$ . The tangent is therefore the horizontal line $y=-16$ .

(3 marks)

Tier 2 · Standard

Find and classify all stationary points of

f(x)=x^3-3x^2-9x+5

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Local maximum $(-1,10)$ Local minimum $(3,-22)$	5
Notes
$f'(x)=3x^2-6x-9=3(x+1)(x-3)$ , so stationary points occur at $x=-1$ and $x=3$ . Their coordinates are $(-1,10)$ and $(3,-22)$ . Since $f''(x)=6x-6$ , $f''(-1)=-12<0$ gives a local maximum and $f''(3)=12>0$ gives a local minimum.

(5 marks)

Tier 3 · Hard

A point

P

lies on the curve

y=\dfrac{12}{x}

, where

x>0

. Use calculus to find the exact coordinates of the point on the curve closest to the origin, and find this minimum distance.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$P=(2\sqrt3,2\sqrt3)$ Minimum distance $2\sqrt6$	6
Notes
The squared distance is $D^2=x^2+y^2=x^2+144x^{-2}$ . Differentiating gives $\dfrac{\mathrm d(D^2)}{\mathrm dx}=2x-288x^{-3}$ . Setting this to zero gives $x^4=144$ , so $x=2\sqrt3$ because $x>0$ . Then $y=12/x=2\sqrt3$ . Also the second derivative is $2+864/x^4>0$ , so this is a minimum. Hence $D=\sqrt{12+12}=2\sqrt6$ .

(6 marks)

Tier 2 · Standard

Find the coordinates of the points on the curve

y=x^3-3x

at which the tangent is parallel to the line

y=9x+2

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$(-2,-2)$ and $(2,2)$	4
Notes
The given line has gradient $9$ . For the curve, $\dfrac{\mathrm dy}{\mathrm dx}=3x^2-3$ . Parallel tangents therefore satisfy $3x^2-3=9$ , so $x^2=4$ and $x=\pm2$ . Substitution into $y=x^3-3x$ gives $y=-2$ when $x=-2$ and $y=2$ when $x=2$ .

(4 marks)

Tier 3 · Hard

A rectangle has vertices

(-x,0)

(x,0)

(-x,12-x^2)

and

(x,12-x^2)

, where

0<x<2\sqrt3

. Use calculus to find the dimensions and maximum area of the rectangle.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Dimensions $4$ by $8$ Maximum area $32$ square units	6
Notes
The width is $2x$ and the height is $12-x^2$ , so $A=2x(12-x^2)=24x-2x^3$ . Then $\dfrac{\mathrm dA}{\mathrm dx}=24-6x^2$ , which is zero at $x=2$ in the stated domain. Also $\dfrac{\mathrm d^2A}{\mathrm dx^2}=-12x<0$ there, so this gives a maximum. The width is $2x=4$ , the height is $12-x^2=8$ , and the maximum area is $4\times8=32$ square units.

(6 marks)

Tier 3 · Hard

For the family of curves

y=x^3-3kx+2

, where

k

is real, determine the number and nature of the stationary points for

k<0

k=0

and

k>0

. Hence find and classify the stationary points when

k=4

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$k<0$ : no stationary points $k=0$ : one stationary point of inflection at $(0,2)$ $k>0$ : a local maximum at $x=-\sqrt{k}$ and a local minimum at $x=\sqrt{k}$ When $k=4$ , local maximum $(-2,18)$ and local minimum $(2,-14)$	7
Notes
$\mathrm dy/\mathrm dx=3(x^2-k)$ . If $k<0$ , this derivative is always positive. If $k=0$ , it is zero only at $x=0$ and does not change sign, giving a stationary point of inflection at $(0,2)$ . If $k>0$ , the roots are $x=\pm\sqrt{k}$ ; the derivative changes from positive to negative at the negative root and from negative to positive at the positive root. For $k=4$ , substitution of $x=-2$ and $x=2$ gives $y=18$ and $y=-14$ respectively.

(7 marks)

Q10

Tier 3 · Hard

10.

Two distinct tangents to the parabola

y=x^2

pass through the point

P(0,-4)

. Find the coordinates of both points of contact and the equations of the two tangents. Hence find an exact expression for the acute angle between the tangents.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Points of contact $(-2,4)$ and $(2,4)$ Tangents $y=-4x-4$ and $y=4x-4$ Acute angle $=\tan^{-1}\!\left(\dfrac8{15}\right)$	6
Notes
At the point $(a,a^2)$ , the parabola has gradient $2a$ , so its tangent is $y-a^2=2a(x-a)$ , or $y=2ax-a^2$ . Passing through $(0,-4)$ requires $-4=-a^2$ , hence $a=\pm2$ . These give the contact points $(-2,4)$ and $(2,4)$ and the two stated tangent equations. If $\theta$ is the acute angle between lines of gradients $4$ and $-4$ , then $\tan\theta=\left\|\dfrac{4-(-4)}{1+(4)(-4)}\right\|=8/15$ . Therefore $\theta=\tan^{-1}(8/15)$ .

(6 marks)

Verified exam appearances

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