The big idea: A graph shows input x on the horizontal axis and output y on the vertical axis.
Each point (x, y) on the graph means y = f(x) for that x.
Graph of y = x + 2. Use Step 1 to mark where the line crosses the y-axis — that is f(0). Step 2 shows how the gradient moves you to the next point. Every point on the line is one (x, y) pair.
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IB language: Read from x to y when asked for f(a).
Find x = a on the horizontal axis, move up or down to the line, then read the y-value.
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The big idea: Plot key points first, then draw smooth or straight connections based on function type.
Example: Sketching a linear function
Sketch y = 2x + 1 for x = −1, 0, 1, 2.
STEPS
- Build a value table.
- Mark y-intercept: the line crosses the y-axis at (0, 1).
- Use the gradient m = 2: go right 1, up 2, to reach (1, 3).
- Plot all four points and draw a straight line through them.
Final answer
A straight line through (−1, −1), (0, 1), (1, 3), (2, 5).
Follow the animation: Step 1 marks the y-intercept at (0, 1). Step 2 uses the gradient m = 2 — go right 1, rise up 2 — to pin the slope. Then the line is drawn through both points.
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The same two-step method works for any linear function.
Try the lines below:
Switch between equations. Notice: a larger |m| makes the line steeper; m = 0 gives a flat horizontal line; c shifts the line up or down.
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Worked example — sketch a quadratic with a vertex
A drone's vertical position z (in m) at horizontal distance x (in m) from launch is modelled by
z(x) = x² − 6x, for 0 ≤ x ≤ 10.
Sketch the graph of y = z(x) for the given domain, clearly showing the vertex and the values at the endpoints.
Step by step
- Step 1 — What shape? The x² is positive, so the curve is a U (smile 😊). The vertex is the lowest point.
- Step 2 — Find the zeros by factoring:
- So the curve passes through (0, 0) and (6, 0).
- Step 3 — Find the vertex and the right endpoint. The vertex sits midway between the zeros (parabola is symmetric): x = (0 + 6)/2 = 3.
- Right endpoint at x = 10:
- Step 4 — Plot the four points and draw the curve. Set up axes that fit all the points (x: 0 to 10, y: about −10 to 40). Plot (0, 0), (3, −9), (6, 0), (10, 40) and draw a smooth U through them.
IB marks (3-mark sketch): • Smooth U-shape in correct window → 1 mark • Vertex (3, −9) labelled → 1 mark • Endpoints (0, 0) and (10, 40) labelled → 1 mark
Final answer
A smooth U-curve from (0, 0) down to the vertex (3, −9), back up through (6, 0), and ending at (10, 40).
Building the sketch step-by-step. Plot the zeros, then the vertex, then the right endpoint — then draw the smooth U.
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IB sketch rule: You only need two points to draw an exact straight line — but plotting a third point is a free check.
If all three are collinear you have no arithmetic error.
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Move 1 — Find f(a): Given an x, find the y.
UP from x, then ACROSS to the y-axis.
Question — find f(3): For the curve y = x² − 2x, find f(3).
Move 1 — Find f(3) on the curve y = x² − 2x. UP from x = 3, then ACROSS to the y-axis.
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Move 2 — Solve f(x) = k: Given a y, find the x.
ACROSS from y, then DOWN to the x-axis.
Question — solve f(x) = 3: For the curve y = x² − 2x, solve f(x) = 3.
Move 2 — Solve f(x) = 3 on the curve y = x² − 2x. ACROSS from y = 3, then DOWN to the x-axis.
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Reading f(a) from a graph
The graph of f is given.
Using the graph, find f(3) and find x when f(x) = 0.
Step by step
- For f(3): start at x = 3 on the x-axis. Go vertically up to the curve.
- The y-value at the curve above x = 3 is 5.
- For f(x) = 0: start at y = 0 (the x-axis) and find where the curve touches it.
- The curve crosses y = 0 at x = −2 and x = 4.
Final answer
f(3) = 5. The function equals zero when x = −2 or x = 4.
Animated walkthrough of the worked example. Watch f(3) = 5 (UP then ACROSS), then x = -2 or 4 (the curve hits the x-axis at two places).
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IB tolerance on graph reading: When reading from a graph (Paper 1), IB usually accepts answers within ±0.2 of the exact value.
If the curve passes through exactly (3, 5), answers of 4.8 to 5.2 are accepted.
Use a ruler and read carefully.
Hover over any point on the graph to read its exact (x, f(x)) coordinates. Practice reading f(a) and finding where f(x) = k.
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The big idea: IB uses the same five function families repeatedly. If you can identify the shape from a graph, you can state the function type instantly — without algebra.
Learn the characteristic shape of each family.
| Family | Typical shape | Key feature to spot |
|---|---|---|
| Linear y = mx + c | Straight line | No curve at all |
| Quadratic y = ax²+bx+c | U-shape (a>0) or ∩-shape (a<0) | One turning point, symmetric |
| Exponential y = abˣ | Rapid growth/decay curve | Curve flattens out — never quite touches one of the axes |
| Power y = axⁿ | Curve through origin or near it | Depends on n: cubic has inflection point |
| Sinusoidal y = a sin(bx)+c | Wave, repeating equally | Regular peaks and troughs, periodic |
Switch between the five function families. Notice the characteristic shape of each. Try to identify the family before revealing the equation.
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Identifying the family from a graph
A graph shows a curve that starts high, decreases, and approaches but never crosses the x-axis.
What function family is it most likely?
Step by step
- Notice the shape: the curve drops steeply at first, then flattens out as it moves right.
- It is not a straight line (rules out linear), has no symmetric U-shape (rules out quadratic), and does not repeat (rules out sinusoidal).
- A smooth decay shape that flattens toward the x-axis is the signature of the exponential family, decay case.
Final answer
Exponential decay: y = abx where 0 < b < 1.
Exponential decay — watch the shape: starts high, drops fast, then flattens toward the x-axis.
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Exam recognition shortcut: In the exam, these four features quickly identify a family:
🔵 No curve at all → linear 🔵 One turning point, symmetric → quadratic 🔵 Curve flattens out, never quite touches an axis → exponential 🔵 Regular waves → sinusoidal
State the family first, then find parameters.