Tables, function graphs and graphical solutions
Plot ordered pairs with suitable scales and use intersections to solve an equation approximately.
Section Paper 1 · Core · Non-calculator
Exam scope and assessment notes
Core: ax + b, ±x² + ax + b and a/x with integer constants. Extended also includes rational coefficients, cubic, reciprocal-square, square-root, inverse-square-root and exponential forms and their sums. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Substitute negative values into expressions with brackets.
- Plot ordered pairs using labelled axes and consistent scales.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is x² − 2 at x = −2?
What you will learn
- Construct tables for linear, quadratic and reciprocal functions.
- Recognise Extended power and exponential graph families and their domains.
- Read roots and line–curve intersections as graphical solutions.
Make the table before drawing
For y = x² − 2, inputs −2, −1, 0, 1 and 2 give outputs 2, −1, −2, −1 and 2. Plot the pairs, including their signs, on labelled axes. Draw a smooth curve through the points. A quadratic is not a chain of straight segments merely because the table has a few sampled inputs.
For y = 4/x, x = 0 is undefined and must not receive a plotted point. Inputs −4, −2, −1, 1, 2 and 4 produce −1, −2, −4, 4, 2 and 1. Plot two separate branches rather than joining through the forbidden input. Use extra points where curvature is stronger and select a scale that shows the requested interval.
Extended: identify the formula's domain and shape
Positive cubic powers x³ allow positive and negative inputs and have an S-shaped basic graph. The reciprocal square 1/x² is positive on both sides of zero and is undefined at zero. The square root √x has real domain x ≥ 0; the inverse square root 1/√x requires x > 0. Negative and fractional powers therefore need domain checks before a table is formed.
An exponential such as y = 2ˣ has outputs 1/4, 1/2, 1, 2 and 4 at x = −2, −1, 0, 1 and 2. It stays positive. A decay model such as y = 80(1/2)ˣ halves for each unit input increase. A shifted or combined expression must be evaluated as a whole, for example y = 2x + 3/x² for x ≠ 0.
An intersection satisfies both expressions
The roots of y = f(x) are the x-coordinates where the curve meets y = 0. To solve f(x) = g(x), draw both graphs on the same axes and read the x-coordinates of their intersections. The y-coordinates are useful for checking the points but are not themselves the solutions for x.
For x² − 2 = x, draw the parabola y = x² − 2 and the line y = x. Their intersections are (−1, −1) and (2, 2), so the solutions are x = −1 and 2. On a hand-drawn graph a reading is generally approximate; state the accuracy that the scale supports. Algebra can check a reading, but use the requested graphical method when the question requires it.
Pause and explain
When solving f(x) = g(x) from two graphs, which coordinates give x?
Worked example
Use a table and a graph to solve x² − 2 = x.
Show the worked solution
- For x = −2, −1, 0, 1, 2, the parabola's y values are 2, −1, −2, −1, 2.
- Plot the smooth parabola and the straight line y = x on the same axes.
- Read the intersections (−1, −1) and (2, 2); their x-coordinates solve the equation.
Answer x = −1 and 2
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Make a table for y = −x² + 4 at x = −2, −1, 0, 1, 2 and state its roots.
Give me a hint
Square the bracketed input before applying the outside negative sign.
Compare my reasoning
- The outputs are 0, 3, 4, 3 and 0.
- Plot a downward-opening smooth curve with symmetry about x = 0.
- It meets the horizontal axis at x = −2 and x = 2.
y values 0, 3, 4, 3, 0; roots −2 and 2
Look for these in your work
- Distinguished −x² from (−x)².
- Read roots from zero outputs.
For y = 6/x, find y at x = −3, −2, 2, 3 and explain why x = 0 cannot be plotted.
Give me a hint
Evaluate each quotient and keep the branches separate.
Compare my reasoning
- The negative inputs give y = −2 and −3.
- The positive inputs give y = 3 and 2.
- At zero the denominator vanishes, so there is no graph point and no line joining the two branches through it.
−2, −3, 3, 2 respectively; undefined at x = 0
Look for these in your work
- Kept input and output signs paired.
- Excluded division by zero.
Extended: compare model A, y = 4x + 4, and model B, y = 2ˣ, at integer x from 0 to 5. Between which consecutive inputs do their graphs switch order?
Give me a hint
Build both tables on the same input values.
Compare my reasoning
- A gives 4, 8, 12, 16, 20, 24; B gives 1, 2, 4, 8, 16, 32.
- At x = 4, A exceeds B; at x = 5, B exceeds A.
- The continuous graphs cross between four and five; the table alone does not give an exact intersection.
Between x = 4 and x = 5
Look for these in your work
- Compared models at matching inputs.
- Separated a bracketed interval from an exact graphical reading.
Common mistakes
- Joining reciprocal branches across zero.
- Drawing a curved function as a polygon without instruction.
- Reporting intersection y values when solving for x.
What can you explain now?
State the roots of y = x² − 9.
Compare with the explanation
x = −3 and 3
The curve has zero output when x² = 9, giving both square-root signs.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Plot a quadratic and a reciprocal using your own tables. Extended: also compare x³, 1/x², √x, 1/√x and 2ˣ, stating each real domain.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.