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O-Level Mathematics · Secondary 3 and 4

Functions and Graphs

Graphs show how one quantity changes with another. In O-Level Mathematics you draw graphs from tables of values, recognise the shapes of linear, quadratic, cubic, reciprocal and exponential graphs, and use a graph to solve equations and estimate gradients. Neat plotting and careful reading of scales matter, because later parts of a question usually use the graph you have drawn. These ideas connect algebra and geometry and prepare you for calculus in Additional Mathematics.

What the syllabus expects

  • Complete a table of values and draw the graph of a given function to a stated scale
  • Recognise and sketch the graphs of y = ax + b, y = ax² + bx + c, y = axⁿ for n = −2, −1, 0, 1, 2 and 3, and y = kaˣ
  • Find the turning point, line of symmetry and intercepts of a quadratic graph
  • Solve equations graphically by drawing a suitable straight line on the same axes
  • Estimate the gradient of a curve at a point by drawing a tangent

Shapes you should recognise

A linear graph y = mx + c is a straight line with gradient m and y-intercept c. A quadratic graph y = ax² + bx + c is a parabola, U-shaped when a is positive and an upside-down U when a is negative, with a line of symmetry through its turning point. A cubic y = ax³ passes through the origin and rises from left to right when a is positive. The reciprocal graph y = a/x has two separate branches and never touches either axis, and for positive a the branches lie in the first and third quadrants. The graph of y = a/x² also never touches the axes, but for positive a both branches lie above the x-axis, because x² is always positive. An exponential graph y = kaˣ with k > 0 and a > 1 passes through (0, k), rises more and more steeply to the right and gets closer and closer to the x-axis on the left without reaching it. A sketch should show the correct shape and any intercepts.

Quadratic graphs in detail

For y = ax² + bx + c, the y-intercept is c, found by putting x = 0. The x-intercepts are the solutions of ax² + bx + c = 0. If the equation factorises as y = (x − p)(x − q), the curve crosses the x-axis at x = p and x = q, and the line of symmetry is halfway between them, at x = (p + q)/2. If the equation is written as y = (x − h)² + k, the turning point is (h, k) and it is a minimum. For y = −(x − h)² + k, the turning point (h, k) is a maximum. For example, y = x² + 2x − 8 = (x + 4)(x − 2) crosses the x-axis at x = −4 and x = 2, so its line of symmetry is x = −1, and substituting x = −1 gives the minimum point (−1, −9). When you draw a parabola, join the points with a smooth curve through the turning point, and make the bottom rounded rather than pointed.

Solving equations with a graph

Once a curve is drawn, you can solve related equations without drawing a new curve. The x-coordinates of the points where a curve meets a straight line are the solutions of the equation formed by setting the two expressions equal. Suppose you have drawn y = x² + x − 6 and want to solve x² + x − 3 = 0. Rearrange so that the left side matches the curve: x² + x − 6 = −3. So draw the horizontal line y = −3 and read the x-coordinates where it crosses the curve. To solve x² − 2 = 0 with the same curve, subtract the expressions: (x² + x − 6) − (x² − 2) = x − 4, so x² + x − 6 = x − 4 and the line to draw is y = x − 4. Read solutions to the accuracy the grid allows, usually 1 decimal place, and check that the number of intersections matches the number of solutions you give.

Gradients of curves

The gradient of a straight line is constant, but the gradient of a curve changes from point to point. To estimate the gradient at a point, draw a tangent: a straight line that just touches the curve at that point. Use a ruler and make the tangent long, then choose two points on it that are far apart and easy to read, and calculate the change in y divided by the change in x. Read the values using the scales on the axes, not by counting squares, because the two axes often have different scales. The gradient is positive where the curve slopes upwards from left to right, negative where it slopes downwards and zero at a turning point. In a real context the gradient has units. On a graph of distance in metres against time in seconds, the gradient of a tangent gives the speed in m/s at that instant.

Worked example

(a) Calculate the values of y = x² − 2x − 3 for x = −2, −1, 0, 1, 2, 3 and 4, and draw the graph. (b) Write down the minimum point and the equation of the line of symmetry. (c) By drawing a suitable straight line, solve x² − 3x − 1 = 0.

  1. Substitute each x value. For example, x = −2 gives 4 + 4 − 3 = 5. The y values for x = −2 to 4 are 5, 0, −3, −4, −3, 0 and 5.
  2. Plot the seven points and join them with a smooth U-shaped curve.
  3. The lowest point is (1, −4), and the curve is symmetrical about the vertical line through it, x = 1.
  4. For part (c), subtract the equations: (x² − 2x − 3) − (x² − 3x − 1) = x − 2.
  5. So x² − 3x − 1 = 0 is the same as x² − 2x − 3 = x − 2. Draw the line y = x − 2, for example through (0, −2) and (4, 2).
  6. Read the x-coordinates of the two intersection points: x ≈ −0.3 and x ≈ 3.3.
  7. Check with the formula: x = (3 ± √13)/2 gives −0.30 and 3.30, which agrees with the graph.

Answer: (b) Minimum point (1, −4), line of symmetry x = 1. (c) Draw y = x − 2, giving x ≈ −0.3 or x ≈ 3.3.

Common mistakes

  • Joining plotted points with straight lines. A quadratic, cubic or reciprocal graph must be a single smooth curve. A ruler is only for straight-line graphs and tangents.
  • Squaring negative x values wrongly on the calculator. Typing −2² gives −4, but in y = x² − 2x − 3 with x = −2 you need (−2)² = 4. Use brackets.
  • Drawing the wrong line for a graphical solution. Rearrange the new equation so that one side is exactly the expression of the curve already drawn, and draw the other side. A subtraction slip here makes every later answer wrong, so expand back to check.
  • Joining the branches of a reciprocal graph. y = 12/x is not defined at x = 0, so the two branches must not be connected across the y-axis.
  • Counting squares to find a gradient. If one square is 0.5 units on the x-axis and 2 units on the y-axis, counting squares gives the wrong answer. Use actual values read from the scales.
  • Drawing a tangent that cuts through the curve near the point. Adjust the ruler until the line just touches the curve at that point, with the curve on one side of it nearby.

Exam tips

  • Plot points with small, neat crosses. If one point sits clearly off the smooth curve, recheck the calculation for that point before you draw.
  • When reading solutions from a graph, draw faint vertical lines from the intersection points down to the x-axis, and give values to the accuracy of the scale.
  • For a tangent, make the line as long as the grid allows and pick two points on it with easy coordinates. A larger triangle gives a more accurate gradient.

Practice questions

1. For the graph of y = 2x² − 8, find the y-intercept and the x-intercepts.

Show answer

The y-intercept is −8, found by putting x = 0. For the x-intercepts, solve 2x² − 8 = 0, so x² = 4 and x = 2 or x = −2.

2. The graph of y = x² + x − 6 has been drawn. Find the equation of the straight line you would draw to solve x² − x − 2 = 0, and state the solutions.

Show answer

Draw y = 2x − 4. Subtracting gives (x² + x − 6) − (x² − x − 2) = 2x − 4. The solutions are x = −1 and x = 2, since x² − x − 2 = (x − 2)(x + 1).

3. The curve y = k × aˣ, where a > 0, passes through the points (0, 3) and (2, 48). Find k and a, then find y when x = 1.5.

Show answer

k = 3, a = 4 and y = 24. At x = 0, y = k × a⁰ = k, so k = 3. At x = 2, 3a² = 48, so a² = 16 and a = 4, since a is positive. When x = 1.5, y = 3 × 4^1.5 = 3 × 8 = 24.

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