Sketching curves by hand is a core requirement for O Level, IGCSE, and A Level exams. You cannot bring graphic display calculators into the exam hall — you must learn to recognise shapes, find asymptotes, trace turning points, and work backwards to find equations. This page teaches you how, with fully worked examples and practice questions.
1. Graph Shapes & Categories
Cambridge syllabi require familiarity with several standard functional shapes. These vary in complexity and are examined across O Level, IGCSE, and A Level courses.
Follow these 4 steps to sketch any quadratic $y = ax^2 + bx + c$ without plotting point by point.
1
Shape — Smiley or Frowny?
Look at the coefficient $a$. If $a > 0$, the curve is U-shaped (minimum). If $a < 0$, it is ∩-shaped (maximum).
2
Find the $y$-intercept
Set $x = 0$. The $y$-intercept is at $(0, c)$. Mark this on your vertical axis.
3
Find the $x$-intercepts (Roots)
Set $y = 0$ and solve $ax^2 + bx + c = 0$ by factorising or the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Check the discriminant $\Delta = b^2 - 4ac$:
If $a > 0$: as $x \to -\infty$, $y \to -\infty$ and as $x \to +\infty$, $y \to +\infty$ (bottom-left to top-right). If $a < 0$: the opposite (top-left to bottom-right).
2
$y$-intercept
Set $x = 0$. The $y$-intercept is $(0, d)$.
3
$x$-intercepts (Roots)
Set $y = 0$ and factorise. Root behaviour depends on multiplicity:
Single root $(x - r)$: Curve crosses straight through the axis.
Repeated root $(x - r)^2$: Curve touches and turns back.
Triple root $(x - r)^3$: Curve has a point of inflection at the axis.
4
Turning Points (A Level)
Differentiate: $\dfrac{dy}{dx} = 3ax^2 + 2bx + c$. Set $\dfrac{dy}{dx} = 0$ and solve for the $x$-coordinates of turning points. Substitute back to find $y$-values.
Cubic — 3 Worked Examples
Example 1: $y = (x - 1)(x + 2)(x - 3)$
Step 1 — End behaviour
Expanding the leading terms: $x \cdot x \cdot x = x^3$, so $a = 1 > 0$. The curve goes from bottom-left to top-right.
Step 2 — $y$-intercept
Set $x = 0$:
$$y = (0 - 1)(0 + 2)(0 - 3) = (-1)(2)(-3) = 6$$
$y$-intercept is $(0, 6)$.
Step 3 — $x$-intercepts
Already factorised. Set each factor to zero:
$$x = 1, \quad x = -2, \quad x = 3$$
Three distinct single roots — the curve crosses the $x$-axis at each one.
Final Sketch
Example 2: $y = -x^3 + 4x$
Step 1 — End behaviour
Leading coefficient is $a = -1 < 0$: top-left to bottom-right.
Trigonometric graphs are periodic — they repeat the same shape over and over. The exam expects you to sketch $\sin$, $\cos$, and $\tan$ from memory and apply transformations. Here's how to learn them systematically.
5a. Key Value Patterns — Learn These by Heart
The trick to drawing $\sin(x)$ and $\cos(x)$ from memory is to memorise their values at the five key angles in one period:
$y = \sin(x)$
00°
→
190°
→
0180°
→
−1270°
→
0360°
$\sin$ starts at zero, goes up to 1, back to zero, down to $-1$, and back to zero.
$y = \cos(x)$
10°
→
090°
→
−1180°
→
0270°
→
1360°
$\cos$ starts at one, drops to zero, down to $-1$, back to zero, up to 1. It's the same wave as $\sin$, shifted left by $90°$.
$y = \tan(x)$
00°
→
∞90°
→
0180°
→
∞270°
→
0360°
$\tan$ has vertical asymptotes at $90°, 270°, \ldots$ where it is undefined. Its period is $180°$ (not $360°$).
5b. Exact Values You Must Memorise
These exact values appear in almost every Cambridge exam. Learn them cold.
Angle
$0°$
$30°$
$45°$
$60°$
$90°$
$\sin\theta$
$0$
$\dfrac{1}{2}$
$\dfrac{\sqrt{2}}{2}$
$\dfrac{\sqrt{3}}{2}$
$1$
$\cos\theta$
$1$
$\dfrac{\sqrt{3}}{2}$
$\dfrac{\sqrt{2}}{2}$
$\dfrac{1}{2}$
$0$
$\tan\theta$
$0$
$\dfrac{1}{\sqrt{3}}$
$1$
$\sqrt{3}$
undefined
Memory trick: For $\sin$, the numerators go $0, 1, \sqrt{2}, \sqrt{3}, 2$ — all divided by $2$. For $\cos$, it's the same sequence in reverse.
5c. The ASTC Rule — "All Students Take Calculus"
Beyond $90°$, you need to know which trig functions are positive in which quadrant:
Quadrant I ($0°$ – $90°$): All are positive
Quadrant II ($90°$ – $180°$): Only Sin is positive
Quadrant III ($180°$ – $270°$): Only Tan is positive
Quadrant IV ($270°$ – $360°$): Only Cos is positive
Use this to find values beyond the first quadrant. For example: $\sin(150°) = \sin(180° - 30°) = +\sin(30°) = \frac{1}{2}$ (Quadrant II, sin positive).
Every exam transformation question boils down to understanding what $a$, $b$, and $c$ do:
$a$
Amplitude
Stretches the wave vertically. The wave oscillates between $c - |a|$ and $c + |a|$. If $a < 0$, the wave is reflected in the $x$-axis.
$$\text{Amplitude} = |a|$$
$b$
Period
Compresses or stretches horizontally. Larger $b$ = shorter period = more waves in the same space.
$$\text{Period} = \frac{360°}{b}$$
$c$
Vertical Shift
Moves the entire wave up (if $c > 0$) or down (if $c < 0$). The midline of the wave is at $y = c$.
$$\text{Midline}: y = c$$
How to sketch $y = a\sin(bx) + c$ step by step
1
Draw the midline
Draw a dashed horizontal line at $y = c$. The wave oscillates around this line.
2
Mark the amplitude
From the midline, mark $|a|$ units up (max = $c + |a|$) and $|a|$ units down (min = $c - |a|$). Draw dashed lines for max and min.
3
Calculate the period and mark key angles
Period = $\frac{360°}{b}$. Divide into 4 equal parts to get the 5 key $x$-values. E.g. if period = $180°$, the key angles are $0°, 45°, 90°, 135°, 180°$.
4
Plot the key values and connect smoothly
For $\sin$: midline → max → midline → min → midline. For $\cos$: max → midline → min → midline → max. If $a < 0$, flip the pattern.
Visual Comparison
See how changing $a$, $b$, and $c$ transforms the original $y = \sin(x)$:
$y = \sin(x)$ (original)
$y = 2\sin(x)$ (amplitude × 2)
$y = \sin(2x)$ (period halved)
$y = \sin(x) + 1$ (shifted up)
Worked Example: $y = 3\sin(2x) - 1$
Identify $a$, $b$, $c$
$a = 3$, $b = 2$, $c = -1$.
Amplitude
$|a| = 3$. The wave goes 3 units above and below the midline.
Period
$\dfrac{360°}{b} = \dfrac{360°}{2} = 180°$. One full wave every $180°$.
Midline & Range
Midline at $y = c = -1$. Maximum = $-1 + 3 = 2$. Minimum = $-1 - 3 = -4$.
Key angles
Period $= 180°$, so quarter-period $= 45°$. The 5 key $x$-values are: $0°, 45°, 90°, 135°, 180°$.
At these angles: $y = -1, \; 2, \; -1, \; -4, \; -1$ (following the sin pattern: midline → max → midline → min → midline).
6. Test Yourself
Try these questions on your own first, then click Reveal Answer to see the full worked solution.
Question 1Quadratic
Sketch $y = x^2 - 6x + 8$, showing all intercepts and the vertex.
