Algebra

Graphs

Year 10 · Year 11

  • Plot and interpret straight-line graphs, including finding the gradient and y-intercept.
  • Plot and recognise the shapes of quadratic, cubic, reciprocal, and exponential graphs.
  • Identify key features of quadratic graphs, such as roots, turning points, and lines of symmetry.
  • Interpret and draw real-life graphs, including distance-time and speed-time graphs.
  • Solve problems by drawing and interpreting graphs.

Key concepts

Straight-line Graphs

A straight-line graph is represented by the equation y = mx + c, where 'm' is the gradient and 'c' is the y-intercept. The gradient describes the steepness and direction of the line. A positive gradient means the line slopes upwards from left to right, a negative gradient means it slopes downwards. The y-intercept is the point where the line crosses the y-axis. Parallel lines have the same gradient. Perpendicular lines have gradients whose product is -1.

y = mx + c
Quadratic Graphs

A quadratic graph is represented by an equation of the form y = ax² + bx + c (where a ≠ 0). Its shape is a parabola. If 'a' is positive, the parabola is U-shaped (minimum point). If 'a' is negative, it's an inverted U-shape (maximum point). Key features include the roots (where the graph crosses the x-axis, i.e., y=0), the turning point (minimum or maximum point), and the line of symmetry (a vertical line passing through the turning point).

y = ax² + bx + c
Cubic Graphs

A cubic graph is represented by an equation of the form y = ax³ + bx² + cx + d (where a ≠ 0). The general shape of a cubic graph has one or two turning points, or a point of inflection where the gradient changes from increasing to decreasing (or vice versa) without a turning point.

y = ax³ + bx² + cx + d
Reciprocal Graphs

A reciprocal graph is represented by an equation of the form y = k/x (where k is a constant). These graphs have two separate branches and never touch the x-axis or y-axis. These axes are called asymptotes.

y = k/x
Exponential Graphs

An exponential graph is represented by an equation of the form y = k^x or y = a^x + b (where k > 0 and k ≠ 1). These graphs show rapid growth or decay. They have a horizontal asymptote, which is a line the graph approaches but never touches. For y = k^x, the x-axis (y=0) is the asymptote. For y = a^x + b, the line y=b is the asymptote.

y = k^x or y = a^x + b
Real-life Graphs

Real-life graphs represent relationships between two quantities in practical contexts. Common examples include distance-time graphs and speed-time graphs. On a distance-time graph, the gradient represents speed. On a speed-time graph, the gradient represents acceleration, and the area under the graph represents the distance travelled. Conversion graphs allow conversion between different units or currencies.

Key facts to remember

  • 1The equation of a straight line is y = mx + c, where m is the gradient and c is the y-intercept.
  • 2Parallel lines have the same gradient. Perpendicular lines have gradients m₁ and m₂ such that m₁m₂ = -1.
  • 3Quadratic graphs (y = ax² + bx + c) are parabolas. 'a' > 0 gives a U-shape (minimum), 'a' < 0 gives an inverted U-shape (maximum).
  • 4The roots of a graph are the x-values where the graph crosses the x-axis (y=0).
  • 5Reciprocal graphs (y = k/x) have asymptotes at x=0 and y=0.
  • 6Exponential graphs (y = k^x) show rapid growth or decay and have a horizontal asymptote.
  • 7On a distance-time graph, the gradient represents speed.
  • 8On a speed-time graph, the gradient represents acceleration, and the area under the graph represents distance travelled.

Worked examples

Example 1

a) Plot the graph of y = 2x - 1 for x values from -2 to 3. b) Find the gradient and y-intercept of the line.

Ia) Create a table of values:
IIx | -2 | -1 | 0 | 1 | 2 | 3
IIIy | -5 | -3 | -1 | 1 | 3 | 5
IVPlot the points (-2, -5), (-1, -3), (0, -1), (1, 1), (2, 3), (3, 5) on a coordinate grid.
VDraw a straight line through the plotted points.
VIb) From the equation y = 2x - 1, compare it to y = mx + c.
VIIThe gradient, m = 2.
VIIIThe y-intercept, c = -1 (or the point (0, -1)).

Answer

a) Graph plotted as described. b) Gradient = 2, y-intercept = -1.

Always use a ruler for straight-line graphs.

Example 2

a) Complete the table of values for y = x² - 2x - 3. b) Plot the graph of y = x² - 2x - 3 for x values from -2 to 4. c) Use your graph to find the roots of the equation x² - 2x - 3 = 0. d) Write down the coordinates of the turning point.

Ia) Calculate y for each x value:
IIx = -2: y = (-2)² - 2(-2) - 3 = 4 + 4 - 3 = 5
IIIx = -1: y = (-1)² - 2(-1) - 3 = 1 + 2 - 3 = 0
IVx = 0: y = (0)² - 2(0) - 3 = -3
Vx = 1: y = (1)² - 2(1) - 3 = 1 - 2 - 3 = -4
VIx = 2: y = (2)² - 2(2) - 3 = 4 - 4 - 3 = -3
VIIx = 3: y = (3)² - 2(3) - 3 = 9 - 6 - 3 = 0
VIIIx = 4: y = (4)² - 2(4) - 3 = 16 - 8 - 3 = 5
9Table of values:
10x | -2 | -1 | 0 | 1 | 2 | 3 | 4
11y | 5 | 0 | -3 | -4 | -3 | 0 | 5
12b) Plot the points on a coordinate grid and draw a smooth curve through them.
13c) The roots are where the graph crosses the x-axis (y=0). From the graph, x = -1 and x = 3.
14d) The turning point is the lowest point on the U-shaped curve. From the graph, this is (1, -4).

Answer

a) Table completed. b) Graph plotted. c) Roots: x = -1, x = 3. d) Turning point: (1, -4).

Ensure your curve is smooth and passes through all plotted points.

Example 3

The graph shows a journey. Section A: (0,0) to (1, 60) - straight line. Section B: (1, 60) to (2, 60) - horizontal line. Section C: (2, 60) to (3.5, 0) - straight line downwards. Describe what is happening in each section (A, B, C) and calculate the average speed for the entire journey.

ISection A: The line is straight and sloping upwards, indicating constant speed. Distance covered = 60 km, Time taken = 1 hour. Speed = Distance / Time = 60 / 1 = 60 km/h.
IISection B: The line is horizontal, meaning the distance from the start is not changing. This represents a stop or a period of no movement. Duration = 2 - 1 = 1 hour.
IIISection C: The line is straight and sloping downwards, indicating constant speed returning to the start. Distance covered = 60 km, Time taken = 3.5 - 2 = 1.5 hours. Speed = Distance / Time = 60 / 1.5 = 40 km/h.
IVTotal distance travelled = 60 km (out) + 60 km (back) = 120 km.
VTotal time taken = 3.5 hours.
VIAverage speed = Total distance / Total time = 120 / 3.5 = 34.285... km/h.
VIIAverage speed ≈ 34.3 km/h (to 3 significant figures).

Answer

Section A: Travelling at a constant speed of 60 km/h for 1 hour. Section B: Stopped for 1 hour. Section C: Returning to the start at a constant speed of 40 km/h for 1.5 hours. Average speed for the entire journey ≈ 34.3 km/h.

Remember that speed is the magnitude of the gradient on a distance-time graph. A negative gradient means returning towards the origin.

Common mistakes

  • Confusing the gradient and y-intercept in y = mx + c.
  • Drawing straight lines for quadratic or cubic graphs instead of smooth curves.
  • Not using a consistent scale on both axes or labelling axes correctly.
  • Misinterpreting the meaning of a horizontal line on a distance-time graph (it means stationary, not constant speed).
  • Incorrectly calculating the gradient, especially with negative coordinates.

Exam tips

  • Always use a sharp pencil and a ruler for plotting graphs.
  • Label your axes clearly and use a consistent scale.
  • For quadratic and cubic graphs, plot enough points to ensure you can draw a smooth, accurate curve.
  • Read the question carefully to determine the required range of x-values and the level of accuracy for answers.
  • When interpreting real-life graphs, pay close attention to the units on the axes.

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