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GRAPHS · LESSON 12

Graphs Mixed Review

Choose the correct graph idea and method without being told what to use.

⏱️ 25–35 min 📘 Review ✏️ Worked examples included

Mixed Review: Choose the Method Yourself

In an examination, the question may not tell you which graph idea to use. Read the information, identify the graph type and decide which relationship matters.

Ask: What do the axes represent? Am I finding a coordinate, gradient, intercept, root, area or a value from a function?

Part A: Coordinates and Scales

  1. State the quadrant containing A(−5, 3).
  2. Point B lies on the x-axis and 7 units left of the origin. Write B.
  3. Choose a sensible scale for values from 0 to 120 using 12 large squares.
  4. Plot P(−2, 4), Q(3, 4), R(3, −1) and S(−2, −1). Name the shape.

Part B: Functions and Straight Lines

  1. Given f(x) = 3x − 2, find f(5).
  2. Given f(x) = 3x − 2 and f(x) = 16, find x.
  3. Find the gradient of the line through (−1, 2) and (3, 10).
  4. Write the equation of a line with gradient −2 and y-intercept 6.
  5. Find the x-intercept of y = 4x − 12.

Part C: Curved Graphs

  1. Complete a table for y = x² − 5 using x = −3, −2, −1, 0, 1, 2, 3.
  2. Draw the graph and estimate its roots.
  3. State why x = 0 cannot be used in y = 3/x.
  4. Describe the basic shape of y = x³.

Part D: Travel Graphs

  1. A person covers 6 km in 30 minutes. Find the speed in km/h.
  2. Explain what a horizontal section means on a distance–time graph.
  3. A car accelerates from 4 m/s to 20 m/s in 8 s. Find acceleration.
  4. Find the distance represented by a speed–time rectangle 12 s wide and 15 m/s high.
  5. A velocity–time graph has +40 m area above the axis and −15 m area below. Find displacement.

Challenge Problems

  1. A line passes through (2, 7) and (6, 19). Find its equation.
  2. Use the graph of y = x² − 4x − 1 to solve x² − 4x − 1 = 3.
  3. A vehicle accelerates uniformly from rest to 18 m/s in 6 s, travels at this speed for 10 s and then decelerates uniformly to rest in 4 s. Find total distance.
  4. Explain why the final position of an object cannot be determined from total distance alone.

✅ Check Your Work

Attempt the questions before opening the solutions.

Show answers and working

Part A

  1. Quadrant II.
  2. B=(−7,0).
  3. 10 units per large square.
  4. A rectangle.

Part B

  1. f(5)=13.
  2. 3x−2=16, so x=6.
  3. Gradient=(10−2)/(3−(−1))=2.
  4. y=−2x+6.
  5. x=3.

Part C

  1. For x=−3 to 3, y=4,−1,−4,−5,−4,−1,4.
  2. Roots are approximately x=−2.24 and x=2.24.
  3. Division by zero is undefined.
  4. An increasing S-shaped curve through the origin.

Part D

  1. 6 km in 0.5 h gives 12 km/h.
  2. The object is stationary.
  3. a=(20−4)÷8=2 m/s².
  4. 12×15=180 m.
  5. 40−15=25 m.

Challenge

  1. Gradient=3, so y=3x+1.
  2. Set y=3 and read intersections; algebraically x²−4x−4=0, so x≈−0.83 or 4.83.
  3. Distance=½×6×18 + 10×18 + ½×4×18 = 54+180+36=270 m.
  4. Distance contains no direction information, so different journeys can have the same distance but different final positions.

Reflection

Revision Guidance and Readiness