Cambridge O Level Mathematics: Graphs of Functions and Sketching Curves — Revision Notes

Condensed for the final weeks. For the full explanation, use the Graphs of Functions and Sketching Curves study guide.

Recognise the shape from the equation

Equation	Shape	Key features
y = mx + c	Straight line	Gradient m, y-intercept c
y = ax² + bx + c	Parabola	Opens up if a > 0, down if a < 0; one turning point
y = ax³ + …	Cubic	Two turning points, or none; opposite ends go opposite ways
y = k/x	Hyperbola	Two branches; asymptotes at both axes
y = ka^x	Exponential	Rapid growth or decay; never reaches the x-axis

The highest power tells you the shape before you plot a single point. That single habit prevents most sketching errors.

Exponential growth and decay

Graphs of the form y = ab^x + c show exponential growth when b > 1, and exponential decay when 0 < b < 1. Typical real-world contexts: population growth and compound interest (growth); radioactive decay and depreciation (decay).

The curve approaches, but never quite reaches, a horizontal asymptote at y = c as x becomes very negative (growth) or very large (decay) — when c = 0, this is simply the x-axis, but a shifted exponential can level off at any horizontal line.

Straight lines

gradient  m = (y2 - y1) / (x2 - x1)
parallel:      same gradient
perpendicular: m1 x m2 = -1

To find the equation from two points: calculate the gradient, then substitute one point into y = mx + c to find c.

Quadratics

Roots are where the curve crosses the x-axis — found by factorising or by the formula.

Completed square form a(x + p)² + q gives the turning point at (−p, q) immediately, and the line of symmetry at x = −p. Alternatively the line of symmetry lies exactly halfway between the two roots.

(Beyond this syllabus, which does not use the term “discriminant”:) the value b² − 4ac tells you how many times the curve meets the x-axis: two, one (tangent), or none.

Key points to mark when sketching

1. y-intercept — set x = 0.
2. x-intercepts (roots) — set y = 0.
3. Turning points.
4. Asymptotes, for reciprocal and exponential graphs.

A sketch does not need to be to scale, but every one of these features must be shown and labelled. That is what the marks are for.

Solving equations graphically

To solve f(x) = g(x), draw both graphs and read the x-coordinates of the intersection points.

To solve f(x) = k, draw the horizontal line y = k and read off where it meets the curve.

Often a question gives you a drawn curve and asks you to solve a different equation. Rearrange the new equation so one side matches the drawn curve — whatever remains is the line you must add.

The answer is always the x-coordinate, not the coordinate pair.

Worked example. To solve x³ + x − 4 = 0 graphically, plot y = x³ + x − 4 and read off the x-value where the curve crosses the x-axis — that x-value is the root of the equation, since setting y = 0 recovers the original equation exactly.

Gradient of a curve

The gradient changes at every point, so it is found by drawing a tangent at the point and calculating that tangent’s gradient.

In context, the gradient is a rate of change:

• Distance–time graph → gradient is speed.
• Speed–time graph → gradient is acceleration, and the area under the graph is the distance travelled. This syllabus only requires areas made up of linear sections (e.g. triangles and trapezia), not curved regions.

Sketching from a table of values vs recognising the equation

Two different exam demands look similar but need different approaches:

• Given an equation, sketch the graph – use the shape-recognition table above first, then mark the key points (intercepts, turning points, asymptotes) rather than plotting many individual points.
• Given a table of values, plot the graph – plot each point accurately on the grid provided, then join with a smooth curve (never straight segments, unless the function genuinely is linear over that interval).

Worked example. A table gives values of y = x² − 2x − 3 for x = −2 to 4. At x = −1, y = (−1)² − 2(−1) − 3 = 1 + 2 − 3 = 0, so (−1, 0) is a root. At x = 3, y = 9 − 6 − 3 = 0, so (3, 0) is the other root. The turning point lies on the line of symmetry, halfway between the roots at x = 1, giving y = 1 − 2 − 3 = −4, so the minimum is (1, −4).

Transformations of graphs

Recognise how shifting or reflecting an equation moves its graph, without needing to re-derive the shape from scratch each time:

Change to equation	Effect on graph
f(x) + a	Shifts up by a (down if a is negative)
f(x + a)	Shifts left by a (right if a is negative)
-f(x)	Reflects in the x-axis

This lets you sketch, for example, y = x² + 3 directly from the standard parabola shape shifted up 3 units, without plotting a single point.

Exam traps

• Plotting points without recognising the expected shape, so an error goes unnoticed.
• Giving the intersection point instead of the x-coordinate.
• Joining points with straight segments where a smooth curve is required.
• Forgetting asymptotes on reciprocal and exponential graphs.
• Reading the gradient of a curve as though it were constant.
• Confusing gradient with area on a speed–time graph.

Self-test

1. What shape is y = 5/x, and where are its asymptotes?
2. Give the condition for two lines to be perpendicular.
3. How do you find the gradient of a curve at a point?
4. On a speed–time graph, what do the gradient and the area under the graph represent?

Answers: 1. A hyperbola with two branches; asymptotes along both the x-axis and the y-axis. 2. The product of their gradients is −1. 3. Draw a tangent to the curve at that point and calculate the gradient of the tangent. 4. The gradient is the acceleration; the area under the graph is the distance travelled.