Chapter 29 – Sketching curves

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Cambridge IGCSE & O Level Mathematics
Chapter 29

Sketching curves

Recognise graph families and produce accurate sketches by labelling intercepts, turning points and asymptotes. Follow the textbook sequence, from straight lines to exponential curves.

Families of graphsLinear graphsQuadratic graphsCubic graphsReciprocal graphsExponential graphs

What you will learn

Exam method: Follow the worked examples from foundation through exam-style. Attempt each question before opening its solution; use your own working to compare steps.

1

Textbook subtopic 1

Families of graphs

Graphs can be grouped by their equations. A linear function makes a straight line; a quadratic makes a parabola; a cubic commonly has an S-shaped or reversed-S-shaped curve; a reciprocal graph has separated branches; and an exponential graph changes by a constant factor for every unit change in \(x\).

To sketch, first recognise the family and the sign of the leading coefficient. Then find useful points by setting \(x=0\) (the vertical-axis intercept) and \(y=0\) (the horizontal-axis intercept). Label any turning points and asymptotes where appropriate. A sketch needs the correct features, not a carefully numbered scale.

Quick identification

\(y=mx+c\): line; \(y=ax^2+bx+c\): parabola; \(y=ax^3+\dots\): cubic; \(y=k/x\): reciprocal; \(y=a^x\), \(a>0, a\ne1\): exponential.

Common exam mistake

The horizontal- and vertical-axis intercepts are different questions. For a horizontal-axis intercept, set y equal to zero, not x.

Worked example 1 · Foundation

Read the degree

Classify \(y=4x^2-3\) and describe its shape.

  1. The largest power of \(x\) is 2.
  2. This is a quadratic; the coefficient of \(x^2\) is positive, so it opens upwards.
Worked example 2 · Developing

Compare two non-linear families

Identify \(y=5/x\) and \(y=2^x\).

  1. \(5/x\) has the variable in the denominator, so it is reciprocal.
  2. \(2^x\) has the variable in the exponent, so it is exponential.
Worked example 3 · Exam-style

Recognise features before sketching

Classify \(y=-x^3+2\) and identify its vertical-axis intercept.

  1. The highest power is 3, so the graph is cubic.
  2. Set \(x=0\) to obtain \(y=2\). The intercept is \((0,2)\).
  3. Because the leading coefficient is negative, the right-hand end of the curve falls.

Practice: Families of graphs

Begin with the basic questions, then attempt the more demanding exam-style items. Open an answer only after trying the question.

Question 1 (Foundation): What is the graph family of \(y=7-2x\)?

Show Answer
  1. It contains only a first power of the variable.
  2. Therefore the graph is a straight line (linear).

Question 2 (Foundation): Classify \(y=6/x\) and name its basic shape.

Show Answer
  1. The variable appears in the denominator.
  2. It is reciprocal: two separate branches approaching the axes.

Question 3 (Developing): Find the vertical-axis intercept of \(y=x^2-5x+8\).

Show Answer
  1. Put \(x=0\).
  2. \(y=0-0+8=8\); the point is \((0,8)\).

Question 4 (Developing): Find the horizontal-axis intercept of \(y=3x-12\).

Show Answer
  1. Set \(y=0\): \(3x-12=0\).
  2. \(x=4\), giving \((4,0)\).

Question 5 (Exam-style): A sketch has a turning point and opens downwards. Which common family is it most likely to be?

Show Answer
  1. A parabola has one turning point.
  2. A downward-opening parabola is quadratic with a negative leading coefficient.

Question 6 (Exam-style): An unshifted graph approaches the two coordinate axes but never meets either axis. Suggest its family and explain.

Show Answer
  1. The two axes act as asymptotes.
  2. A typical reciprocal function \(y=k/x\) with non-zero \(k\) has this behaviour.
2

Textbook subtopic 2

Linear graphs

In \(y=mx+c\), \(m\) is the gradient and \(c\) is the vertical-axis intercept. A positive gradient rises to the right; a negative gradient falls to the right. Two accurate points are sufficient to fix a line.

For a simple sketch, find the intercepts: substitute \(x=0\) for the \(y\)-intercept and \(y=0\) for the \(x\)-intercept. If the line passes through the origin, use one further point.

Common exam mistake

For a horizontal line such as \(y=4\), every point has vertical coordinate 4; there is no horizontal-axis intercept.

Textbook line illustrating a negative gradient and two intercepts.
A descending straight-line graph, with intercepts marked Source: Chapter 29 textbook illustration.
Worked example 4 · Foundation

Two intercepts

Sketch \(y=-2x+8\).

  1. Set \(x=0\): the vertical intercept is \((0,8)\).
  2. Set \(y=0\): \(-2x+8=0\), so \(x=4\).
  3. Draw a descending line through \((0,8)\) and \((4,0)\).
Worked example 5 · Developing

A fractional gradient

Find two convenient points for \(y=\frac{3}{2}x-3\).

  1. At \(x=0\), \(y=-3\).
  2. At \(y=0\), solve \(\frac32 x=3\), giving \(x=2\).
  3. The line passes through \((0,-3)\) and \((2,0)\).
Worked example 6 · Exam-style

Straight line from an equation

Sketch the line \(2x+3y=12\), labelling both intercepts.

  1. At \(x=0\): \(3y=12\Rightarrow y=4\).
  2. At \(y=0\): \(2x=12\Rightarrow x=6\).
  3. Connect \((0,4)\) and \((6,0)\) with a straight line.

Practice: Linear graphs

Begin with the basic questions, then attempt the more demanding exam-style items. Open an answer only after trying the question.

Question 7 (Foundation): Find the vertical-axis intercept of \(y=4x+5\).

Show Answer
  1. At \(x=0\), \(y=5\).
  2. Intercept: \((0,5)\).

Question 8 (Foundation): Find the horizontal-axis intercept of \(y=x-7\).

Show Answer
  1. Set \(y=0\): \(x-7=0\).
  2. Intercept: \((7,0)\).

Question 9 (Developing): Does the graph \(y=-5x+2\) rise or fall to the right?

Show Answer
  1. The gradient is \(-5\).
  2. A negative gradient means the line falls to the right.

Question 10 (Developing): Give two plotting points for \(y=2x\).

Show Answer
  1. At \(x=0\), \(y=0\).
  2. At \(x=2\), \(y=4\).
  3. Use \((0,0)\) and \((2,4)\).

Question 11 (Exam-style): Find both intercepts of \(3x-2y=12\).

Show Answer
  1. If \(y=0\), then \(x=4\).
  2. If \(x=0\), then \(-2y=12\Rightarrow y=-6\).
  3. Label \((4,0)\) and \((0,-6)\).

Question 12 (Exam-style): A line has intercepts \((0,6)\) and \((3,0)\). Find its equation.

Show Answer
  1. Gradient \(m=(0-6)/(3-0)=-2\).
  2. At \(x=0\), the intercept is 6.
  3. The equation is \(y=-2x+6\).
3

Textbook subtopic 3

Quadratic graphs

A quadratic graph \(y=ax^2+bx+c\) is a parabola. If \(a>0\) it has a minimum; if \(a<0\) it has a maximum. The graph is symmetric about a vertical line through the turning point. The \(x\)-intercepts are roots of \(ax^2+bx+c=0\).

Find the roots where possible, the \(y\)-intercept, and the turning point. For \(y=ax^2+bx+c\), the axis of symmetry is \(x=-b/(2a)\). Alternatively, write the quadratic as \(y=a(x-h)^2+k\); then the turning point is \((h,k)\).

Common exam mistake

The turning point is not always on the horizontal axis. Label both of its coordinates, including negative values.

Two textbook parabolas showing the effect of the sign of the quadratic coefficient.
Upward and downward parabola shapes Source: Chapter 29 textbook illustration.
Textbook sketch of a parabola with roots and a turning point.
A labelled quadratic turning point Source: Chapter 29 textbook illustration.
Worked example 7 · Foundation

Roots and y-intercept

Find the intercepts of \(y=x^2-5x+6\).

  1. Set \(y=0\): \((x-2)(x-3)=0\).
  2. The roots are \(x=2,3\), giving \((2,0),(3,0)\).
  3. At \(x=0\), \(y=6\); label \((0,6)\).
Worked example 8 · Developing

Minimum point

Find the turning point of \(y=x^2-4x+1\).

  1. Complete the square: \(x^2-4x+1=(x-2)^2-3\).
  2. The minimum point is \((2,-3)\); the axis is \(x=2\).
  3. The parabola opens upwards.
Worked example 9 · Exam-style

Downward-opening parabola

Sketch \(y=-(x-1)^2+9\) with its intercepts and turning point.

  1. The maximum point is \((1,9)\); the axis is \(x=1\).
  2. Set \(y=0\): \((x-1)^2=9\), so \(x=-2,4\).
  3. At \(x=0\), \(y=8\). Draw a downward parabola through the labelled points.

Practice: Quadratic graphs

Begin with the basic questions, then attempt the more demanding exam-style items. Open an answer only after trying the question.

Question 13 (Foundation): Does \(y=3x^2+1\) have a maximum or a minimum?

Show Answer
  1. The coefficient of \(x^2\) is positive.
  2. The parabola has a minimum.

Question 14 (Foundation): Find the vertical-axis intercept of \(y=x^2+4x-2\).

Show Answer
  1. Set \(x=0\).
  2. \(y=-2\), so the intercept is \((0,-2)\).

Question 15 (Developing): Find the roots of \(y=x^2-9\).

Show Answer
  1. Set \(y=0\): \((x-3)(x+3)=0\).
  2. The roots are \(-3,3\).

Question 16 (Developing): Find the axis of symmetry of \(y=x^2-6x+5\).

Show Answer
  1. Here \(a=1,b=-6\); calculate \(-b/(2a)=6/2=3\).
  2. The axis is \(x=3\).

Question 17 (Exam-style): Find the turning point of \(y=x^2+2x-8\).

Show Answer
  1. Complete the square: \(y=(x+1)^2-9\).
  2. The minimum is at \((-1,-9)\).

Question 18 (Exam-style): For \(y=-x^2+4x+5\), find the maximum and both horizontal-axis intercepts.

Show Answer
  1. Complete the square: \(y=-(x-2)^2+9\); maximum at \((2,9)\).
  2. Solve \(-x^2+4x+5=0\): \((x-5)(x+1)=0\).
  3. Horizontal-axis intercepts: \((-1,0)\) and \((5,0)\).
4

Textbook subtopic 4

Cubic graphs

Cubic functions contain an \(x^3\) term. A positive leading coefficient gives a curve whose left end falls and whose right end rises; a negative leading coefficient reverses this. A cubic can cross the horizontal axis one or three distinct times, or touch at a repeated root.

For a useful sketch, factorise to find real roots when possible, evaluate \(y\) at \(x=0\), and make the curve pass through the intercepts in the correct order. Do not assume that every cubic has two turning points.

Common exam mistake

Repeated factors can give a touch rather than a crossing. Check whether a factor is squared.

Textbook graph of a cubic with three real zeros.
A cubic curve crossing the horizontal axis three times Source: Chapter 29 textbook illustration.
Worked example 10 · Foundation

Factorise to find roots

Sketch the intercept structure of \(y=x(x-2)(x+3)\).

  1. Solve \(x(x-2)(x+3)=0\): roots \(-3,0,2\).
  2. The leading term is \(x^3\), so left end down, right end up.
  3. Cross the horizontal axis at \((-3,0),(0,0),(2,0)\).
Worked example 11 · Developing

Translate the standard cubic

Describe \(y=(x-2)^3+1\).

  1. The basic graph \(y=x^3\) is shifted 2 right and 1 up.
  2. Its central flattening point moves from \((0,0)\) to \((2,1)\).
  3. The vertical-axis intercept is \((-8+1)=-7\), so \((0,-7)\).
Worked example 12 · Exam-style

Negative cubic and repeated roots

For \(y=-x(x-1)^2\), describe the behaviour at \(x=0\) and \(x=1\).

  1. The factors show roots at \(x=0\), and a double root at \(x=1\).
  2. The curve crosses at \(x=0\) but touches/turns at \(x=1\).
  3. The leading term is \(-x^3\), so its left end rises and right end falls.

Practice: Cubic graphs

Begin with the basic questions, then attempt the more demanding exam-style items. Open an answer only after trying the question.

Question 19 (Foundation): Is \(y=x^3+2x\) cubic?

Show Answer
  1. The highest index of \(x\) is 3.
  2. Yes; it is a cubic function.

Question 20 (Foundation): Find the vertical-axis intercept of \(y=x^3-4x+7\).

Show Answer
  1. Set \(x=0\); \(y=7\).
  2. The intercept is \((0,7)\).

Question 21 (Developing): Find the horizontal-axis intercepts of \(y=x(x+1)(x-4)\).

Show Answer
  1. Equate each factor to zero.
  2. \(x=0,-1,4\) gives the three intercepts.

Question 22 (Developing): Which end of \(y=-2x^3+x\) rises?

Show Answer
  1. The leading term is \(-2x^3\).
  2. The left end rises and the right end falls.

Question 23 (Exam-style): For \(y=(x+2)^3-1\), find the central flattening point.

Show Answer
  1. Compare with \(y=(x-h)^3+k\).
  2. Here \(h=-2,k=-1\), so it is \((-2,-1)\).

Question 24 (Exam-style): For \(y=x^3-4x\), find and label its real roots and state the end behaviour.

Show Answer
  1. Factorise: \(x(x^2-4)=x(x-2)(x+2)\).
  2. Roots: \(-2,0,2\).
  3. Positive leading coefficient: left end down, right end up.
5

Textbook subtopic 5

Reciprocal graphs

A basic reciprocal graph \(y=k/x\) has two separate branches. The axes \(x=0\) and \(y=0\) are asymptotes: the curve approaches them but does not meet them. For \(k>0\) branches lie in quadrants I and III; for \(k<0\), quadrants II and IV.

A shifted reciprocal function \(y=k/(x-a)+b\) has asymptotes \(x=a\) and \(y=b\). Mark the asymptotes first, then one point in each branch. The graph is undefined where the denominator is zero.

Common exam mistake

An asymptote is generally drawn dashed. Never draw a basic reciprocal curve through its vertical asymptote.

Two branches of the textbook reciprocal curve and its coordinate-axis asymptotes.
Basic positive reciprocal graph Source: Chapter 29 textbook illustration.
Textbook example of a shifted reciprocal graph.
A shifted reciprocal curve and its dashed asymptote Source: Chapter 29 textbook illustration.
Worked example 13 · Foundation

Basic reciprocal

Sketch \(y=6/x\), with the asymptotes.

  1. Denominator zero at \(x=0\); the vertical asymptote is \(x=0\).
  2. As \(x\) grows large, \(6/x\to0\); horizontal asymptote \(y=0\).
  3. Use points \((2,3)\) and \((-2,-3)\) to sketch two branches.
Worked example 14 · Developing

Horizontal and vertical shifts

Find the asymptotes of \(y=3/(x-2)+1\).

  1. The denominator is zero when \(x=2\); vertical asymptote \(x=2\).
  2. The added 1 shifts the horizontal asymptote to \(y=1\).
  3. Use \(x=3\Rightarrow y=4\) as a plotting point.
Worked example 15 · Exam-style

A negative shifted reciprocal

Sketch \(y=-4/(x+1)+2\) using asymptotes and two points.

  1. Asymptotes: \(x=-1\), \(y=2\).
  2. At \(x=0\), \(y=-2\); at \(x=-2\), \(y=6\).
  3. Plot the points on opposite sides of \(x=-1\); each branch tends towards the asymptotes.

Practice: Reciprocal graphs

Begin with the basic questions, then attempt the more demanding exam-style items. Open an answer only after trying the question.

Question 25 (Foundation): Where is \(y=5/x\) undefined?

Show Answer
  1. The denominator cannot equal zero.
  2. It is undefined at \(x=0\).

Question 26 (Foundation): Find the horizontal asymptote of \(y=7/x\).

Show Answer
  1. \(7/x\to0\) as \(|x|\) increases.
  2. The horizontal asymptote is \(y=0\).

Question 27 (Developing): Find the asymptotes of \(y=2/(x-4)+3\).

Show Answer
  1. Equation \(x-4=0\) gives \(x=4\).
  2. The vertical shift gives \(y=3\).
  3. Asymptotes: \(x=4\), \(y=3\).

Question 28 (Developing): Find one point on \(y=-8/x\) in quadrant IV.

Show Answer
  1. Choose \(x=2\).
  2. Then \(y=-8/2=-4\), so \((2,-4)\) works.

Question 29 (Exam-style): Find the vertical-axis intercept of \(y=4/(x+2)-1\).

Show Answer
  1. Substitute \(x=0\): \(y=4/2-1=1\).
  2. The intercept is \((0,1)\).

Question 30 (Exam-style): For \(y=6/(x-3)-2\), give asymptotes and the horizontal-axis intercept.

Show Answer
  1. The asymptotes are \(x=3\), \(y=-2\).
  2. Set \(y=0\): \(6/(x-3)=2\).
  3. Then \(x-3=3\), so \(x=6\); intercept \((6,0)\).
6

Textbook subtopic 6

Exponential graphs

An exponential function such as \(y=2^x\) has the variable in the power. For a base greater than 1 it rises to the right; for a base between 0 and 1 it falls. The standard graph passes through \((0,1)\) and stays above the horizontal axis.

For \(y=a^x+k\), the horizontal asymptote is \(y=k\). Find the vertical intercept by setting \(x=0\), use a second value such as \(x=1\), and sketch the correct increasing/decreasing shape without crossing the asymptote.

Common exam mistake

A horizontal shift of an exponential must be read from its exponent, but a vertical shift changes the horizontal asymptote.

Textbook exponential curve with a dashed horizontal asymptote.
Shifted increasing exponential graph Source: Chapter 29 textbook illustration.
Worked example 16 · Foundation

Standard exponential

Find three points on \(y=2^x\).

  1. For \(x=-1\), \(y=1/2\).
  2. For \(x=0\), \(y=1\); for \(x=2\), \(y=4\).
  3. The curve rises and approaches \(y=0\) on the left.
Worked example 17 · Developing

Translated exponential

Sketch \(y=2^x-3\), stating its asymptote and intercepts.

  1. Horizontal asymptote: \(y=-3\).
  2. At \(x=0\), \(y=1-3=-2\).
  3. At \(y=0\), \(2^x=3\), so \(x=\log_2 3\approx1.585\).
  4. Label \((0,-2)\) and the approximate horizontal-axis intercept.
Worked example 18 · Exam-style

Decreasing exponential

Identify the main sketch features of \(y=4(1/2)^x+1\).

  1. The base \(1/2\) is between 0 and 1, so the curve decreases to the right.
  2. At \(x=0\), \(y=5\); at \(x=1\), \(y=3\).
  3. Horizontal asymptote: \(y=1\); the curve stays above it.

Practice: Exponential graphs

Begin with the basic questions, then attempt the more demanding exam-style items. Open an answer only after trying the question.

Question 31 (Foundation): Find the vertical-axis intercept of \(y=3^x\).

Show Answer
  1. At \(x=0\), \(3^0=1\).
  2. Intercept: \((0,1)\).

Question 32 (Foundation): Is \(y=(1/3)^x\) increasing or decreasing?

Show Answer
  1. The base \(1/3\) is between 0 and 1.
  2. The function decreases as x increases.

Question 33 (Developing): Find the horizontal asymptote of \(y=5^x-4\).

Show Answer
  1. The basic asymptote \(y=0\) has shifted down 4.
  2. The new asymptote is \(y=-4\).

Question 34 (Developing): For \(y=2^x+3\), calculate \(y\) at \(x=0,1\).

Show Answer
  1. \(x=0\Rightarrow y=1+3=4\).
  2. \(x=1\Rightarrow y=2+3=5\).
  3. Points \((0,4),(1,5)\).

Question 35 (Exam-style): Find the horizontal-axis intercept of \(y=2^x-8\).

Show Answer
  1. Set \(0=2^x-8\): \(2^x=8=2^3\).
  2. \(x=3\); intercept \((3,0)\).

Question 36 (Exam-style): Sketch \(y=4-2^x\), labelling the asymptote and intercepts.

Show Answer
  1. The curve decreases because of the negative sign before \(2^x\).
  2. As \(x\to-\infty\), \(2^x\to0\), so asymptote \(y=4\).
  3. At \(x=0\), \(y=3\); at \(y=0\), \(2^x=4\Rightarrow x=2\).
  4. The intercepts are \((0,3),(2,0)\).
Source and originality: The main headings and mathematical concepts follow the supplied Cambridge IGCSE & O Level Mathematics second-edition textbook. The cropped diagrams are from the corresponding textbook chapter. The explanations, worked examples and exercises are original teaching material; “exam-style” means practice written in an examination-like format, not copied past-paper questions.
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