Functions and Graphs covers the Cartesian plane, straight-line graphs y = ax + b, quadratic graphs y = ax² + bx + c, power graphs y = axⁿ, exponential graphs y = kaˣ, and estimating the gradient of a curve by drawing a tangent. Reading and plotting points on the Cartesian plane underpins all of it and appears in almost every O-Level Paper 1 and Paper 2.
Quadratic graphs are the most heavily examined graphs in Sec 3 E-Math. Paper 1 tests quick sketching from the factorised or completed-square form, and Paper 2 almost always carries a full plot-and-read graph question. Each property of the graph is worth a mark on its own: shape, turning point, symmetry and intercepts.
What these notes cover
- •The Cartesian plane and quadrants
- •Graphs and functions
- •Reading coordinates off a plotted line
- •Linear functions y = ax + b
- •Gradient and y-intercept in context
- •Finding the equation of a line from its graph
- •Quadratic functions and the parabola
- •Sketching from vertex and intercept forms
- •Reading solutions off a graph
- •Power graphs and exponential graphs
- •Estimating the gradient of a curve
- •Common mistakes and exam technique
Key definitions
- Function
- A relationship in which each x-value gives exactly one y-value. The rule “y is a number whose square is x” is not a function, because x = 9 gives both y = 3 and y = −3.
- Gradient
- How steep a line is, measured as vertical change ÷ horizontal change (rise over run). In y = ax + b the gradient is a. A vertical line x = k has an undefined gradient.
- Parabola
- The graph of every quadratic function. It is a single smooth curve with exactly one turning point and is symmetric about a vertical line through that turning point.
- Asymptote
- A line that a curve gets arbitrarily close to but never touches or crosses. Both axes are asymptotes of y = a ÷ x and y = a ÷ x², and the x-axis is an asymptote of y = kaˣ.
- Tangent
- A straight line that touches a curve at one point only. The gradient of a curve at a point is defined as the gradient of the tangent drawn at that point.
Key formulas
Linear function
y = ax + b
a is the gradient, or rate of change. b is the y-intercept: the starting value, where the line cuts the y-axis. Vertical lines x = k cannot be written in this form.
Gradient through two points
Applies to a line through (x₁, y₁) and (x₂, y₂). Keep the two points in the same order on the top and the bottom, or the sign flips.
Quadratic function
y = ax² + bx + c, a ≠ 0
If a > 0 the curve opens upward and has a minimum point. If a < 0 it opens downward and has a maximum point. The curve always cuts the y-axis at (0, c).
Vertex (completed-square) form
y = ±(x − p)² + q
The turning point is (p, q) and the line of symmetry is x = p. A plus sign in front gives a minimum point; a minus sign gives a maximum point.
Line of symmetry from the roots
Applies to the intercept form y = ±(x − a)(x − b), whose roots are x = a and x = b. Here a and b are the roots, not the coefficients from y = ax² + bx + c.
Exponential function
a is a positive integer and k > 0. The graph lies entirely above the x-axis and cuts the y-axis at k. The x-axis is an asymptote.
Worked example
Water in a tank is heated at a constant rate, so its temperature T °C after x minutes follows a linear function T = ax + b. After 2 minutes the temperature is 34 °C, and after 8 minutes it is 58 °C. (a) Find the gradient a. (b) Find b, and interpret a and b in context. (c) Find the time taken for the water to reach 70 °C. (d) Explain why the point (5, 48) does not lie on the line.
- 1(a) The graph passes through (2, 34) and (8, 58), so a = (58 − 34) ÷ (8 − 2) = 24 ÷ 6 = 4.
- 2(b) Substitute a = 4 and the point (2, 34) into T = ax + b: 34 = 4(2) + b, so 34 = 8 + b and b = 26. This gives T = 4x + 26.
- 3In context, b = 26 means the water started at 26 °C, its temperature at x = 0 before heating began. a = 4 means the water heats up at a rate of 4 °C per minute.
- 4(c) Set T = 70: 70 = 4x + 26, so 4x = 70 − 26 = 44 and x = 11. The water reaches 70 °C after 11 minutes.
- 5(d) Substitute x = 5 into the equation: T = 4(5) + 26 = 46.
- 6Since 46 ≠ 48, the point (5, 48) does not satisfy T = 4x + 26, so it does not lie on the line.
Watch out: To show whether a point lies on a line, substitute and compare. A bare “it is not on the graph” scores nothing.
Common mistakes
Swapping coordinate order in the gradient formula
Wrong: With A(1, 5) and B(4, −1), computing (5 − (−1)) ÷ (4 − 1) = 2. This takes the y-values in one order and the x-values in the other, so the size is right but the sign is wrong.
Right: Keep the same point first on the top and the bottom: m = (−1 − 5) ÷ (4 − 1) = −6 ÷ 3 = −2. As a check, a line that falls from left to right must have a negative gradient.
Misreading the sign in completed-square form
Wrong: Given y = (x − 3)² + 2, reading off the turning point as (−3, 2) by copying the sign inside the bracket.
Right: The turning point is where the bracket equals zero. Here x − 3 = 0, so x = 3 and the minimum point is (3, 2). The x-coordinate takes the opposite sign to the one in the bracket, while the number outside keeps its own sign.
Drawing a chord instead of a tangent
Wrong: Joining two points on the curve with a straight line (a chord) and finding its gradient. This gives only an average gradient, not the gradient at the point.
Right: Place the ruler so it touches the curve at that one point only and draw a long tangent, at least 6 cm. Pick two far-apart points on the tangent line, then divide the vertical change by the horizontal change.
Questions students ask
How do you find the turning point of a quadratic graph?
In vertex form y = ±(x − p)² + q, the turning point is (p, q). It is a minimum point with a plus sign in front and a maximum point with a minus sign. In factorised form, the line of symmetry lies halfway between the roots, x = (a + b) ÷ 2, and substituting that x-value back gives the y-coordinate of the turning point.
How do you estimate the gradient of a curve at a point?
Draw a long tangent that touches the curve at that point only. Read two well-separated points on the tangent line, not on the curve, using the axis scales. Divide the vertical change by the horizontal change; examiners accept any answer within a tolerance band.
How do you solve an equation using a graph in E-Math?
Rewrite the equation as (curve) = (number); for example, x² − 2x − 3 = 2 becomes the curve y = x² − 2x − 3 and the line y = 2. Rule the horizontal line y = 2 across your graph and drop vertical dashed lines from both points where it cuts the curve, reading off both x-values. As a check, the midpoint of the two readings should equal the line of symmetry.
What do the gradient and y-intercept mean in a word problem?
In y = ax + b, b is the starting value (the value of y when x = 0) and a is the rate of change. For a courier charging y = 4x + 20 dollars for x parcels, 20 is the booking fee and 4 is the charge per parcel. Always answer in the context of the question, not just “it is the y-intercept”.
In the full PDF
- •All nine figures, including the plotted lines, labelled parabolas and power-graph sketches
- •Two further worked examples: sketching y = −(x − 1)(x − 5) and estimating a gradient from a tangent
- •Quadratics in real contexts: projectile height and maximising the area of a fenced pen
- •The full power-graph gallery for n = −2 to 3 and the y = x³ − 2x table of values
- •Three further common mistakes and the complete exam technique checklist
- •One-page summary of the graph families and how to choose the right quadratic form