Secondary Mathematics: Graphs, Tables and Relationships

Contents · choose your starting point

1. 1. A coordinate pair is an ordered instruction
2. 2. Read the scale before counting squares
3. 3. A table supplies coordinate pairs
4. 4. Constant differences reveal a candidate rate
5. 5. Gradient measures change in y per change in x
6. 6. The intercept tells us the output at zero input
7. 7. Find a straight-line equation from information
8. 8. Direct proportion is a special straight-line relationship
9. 9. An intersection solves two relationships at once
10. 10. Graphs can solve equations with a target level
11. 11. A quadratic is not a straight line between a few dots
12. 12. Extension: reciprocal relationships have excluded inputs
13. 13. A journey graph needs a precise vertical quantity
14. 14. Piecewise rules describe different behaviour on different intervals
15. 15. Data points, fitted models and certainty are different things
16. 16. Capstone: compare two invented pricing rules
17. 17. Independent practice
18. 18. Worked answers
19. 19. Diagnose a graph mistake by its first cause
20. 20. Continue through the BTT Mathematics library
21. Sources and scope

Secondary Mathematics · Worked Repair Guide 04

A graph is a representation of a relationship between quantities. To read it well, identify what each axis measures, what its scale means, which inputs are permitted, and how the plotted points connect to a table or equation. The picture is useful only while those meanings remain attached to it.

This guide teaches a route from coordinates and tables to straight-line equations, gradients, intercepts, intersections, selected curves and real-world models. Its central question is “What does this point or change mean in the original quantities?” The answer should remain consistent whether the relationship is written in words, numbers, symbols or a graph.

The examples are original teaching situations. Numerical tables represent stipulated relationships or invented observations; they are not actual transport measurements, prices or student results. The guide includes instructions for making sketches on squared paper, together with text and tables that remain useful without an interactive graphing tool.

Start with axis reading and straight lines. The curve and modelling sections are extensions to be matched to the learner’s current course rather than treated as one universal secondary-school syllabus sequence.

1. A coordinate pair is an ordered instruction

The point (3, 5) places the first coordinate on the horizontal x-axis and the second on the vertical y-axis. From the origin, move three x-units horizontally and five y-units vertically. The point (5, 3) generally occupies a different position because the order has changed.

For (−2, 4), move left by two x-units and up by four y-units. For (−2, −4), move left by two and down by four. Negative coordinates are positions relative to the chosen origin; they are not automatically impossible values in every context.

A point on the y-axis has x = 0. A point on the x-axis has y = 0. This makes intercepts easy to identify later. The origin is (0, 0), where both coordinates are zero.

If the axes describe time in minutes and distance in metres, the point (3, 120) means 120 metres at three minutes. Reversing the coordinates would mean something quite different. Labels and units are therefore part of the representation, not finishing touches to add after plotting.

Entry check: explain (0, 7), (−3, 0) and (2, −5) in coordinate language. Then consider a model in which x counts purchased notebooks. Would x = 2.5 be permitted? Not for a count of whole notebooks under that definition. A coordinate can be valid on the mathematical plane but outside the model’s allowed inputs.

2. Read the scale before counting squares

A small square does not always represent one unit. If horizontal tick marks are labelled 0, 5, 10 and 15 at equal spacing, each labelled interval represents five x-units. The vertical axis can use a completely different scale.

Suppose a line moves three horizontal squares and four vertical squares. If each horizontal square represents two minutes and each vertical square represents five metres, the changes are six minutes and twenty metres. The gradient in the original quantities is 20/6 = 10/3 metres per minute, not 4/3.

The visual angle depends on how the axes have been drawn. Stretching the vertical scale can make the same relationship appear steeper without changing its numerical gradient in the original units. Comparing the apparent steepness of two separately scaled graphs is therefore unsafe unless their axes are accounted for.

When sketching, choose regular scales that allow the important features to be read. Mark at least enough tick values to make the scale unambiguous. Include zero when needed to interpret intercepts; when a graph uses a restricted display window, make that window clear rather than pretending the visible corner is the origin.

Repair task: draw the same relationship y = 2x on two sets of axes, using different vertical scales. The pictures will look different, but the coordinate pairs and numerical gradient remain the same. Explain which features belong to the relationship and which belong only to the drawing.

3. A table supplies coordinate pairs

For the equation y = 2x − 3, choose several x-values and calculate the corresponding y-values. Each row gives a point on the graph.

At x = −2, calculate 2(−2) − 3 = −7. At x = 0, the output is −3. Plot the resulting points with their coordinates intact, then draw the straight line described by the equation.

The equation, not merely the number of plotted points, tells us that this is a straight-line relationship. A table can help test whether a proposed rule agrees with supplied values, but a finite set of values does not by itself prove that no other rule could fit between them.

For example, the values at x = 0, 1 and 2 can agree with y = 2x − 3, yet the different rule y = 2x − 3 + x(x − 1)(x − 2) also agrees at those three inputs. The additional product is zero there. Away from them, the two rules can differ.

This distinction matters when moving from exact mathematics to data modelling. If an exact equation is given, its graph follows that rule. If only observations are given, a line may be a proposed model rather than a proven law. Keep the source of the relationship clear.

4. Constant differences reveal a candidate rate

In the table for y = 2x − 3, increasing x by one increases y by two. For this given linear rule, the rate of change is constant. If x increases by three, y increases by six. If x decreases by two, y decreases by four.

For a straight line y = mx + c, changing x by an amount Δx changes y by mΔx. The difference between the two outputs is:

[m(x + Δx) + c] − [mx + c] = mΔx.

The constant c cancels. Dividing by a non-zero Δx gives m. This derivation explains why the gradient does not depend on which two distinct points on the line we choose.

A table with unequal x-steps requires division by the actual step sizes. Consider x-values 0, 2 and 5 with corresponding y-values 1, 7 and 16. The y-changes are 6 and 9, while the x-changes are 2 and 3. Both rates equal 3, so the points are consistent with y = 3x + 1.

Looking only for equal y-differences would miss that consistency because the input intervals differ. Conversely, equal output differences over unequal input intervals do not establish a constant rate.

Worked check: if x changes from 2 to 6 and y changes from 9 to 21, the rate over that interval is (21 − 9)/(6 − 2) = 12/4 = 3. Do not divide 21 by 6 unless you are specifically comparing the output with the input from zero in a proportional relationship.

5. Gradient measures change in y per change in x

For two distinct points on a non-vertical straight line:

gradient = (y₂ − y₁)/(x₂ − x₁).

Use the same order in the numerator and denominator. Reversing both orders gives the same quotient; reversing only one changes its sign incorrectly.

Worked example: find the gradient through (−1, 5) and (3, −3). The change in y is −3 − 5 = −8, and the change in x is 3 − (−1) = 4. The gradient is −2.

The negative sign means y decreases as x increases. Specifically, each increase of one x-unit decreases y by two units along this line. It does not mean the line’s y-values are all negative. At x = −1, the y-value is positive 5.

A horizontal line has zero gradient because y does not change when x changes. The line y = 4 is an example. A vertical line such as x = −2 has undefined gradient in this formula because its horizontal change is zero. It is not a line with gradient zero.

When the axes have units, the gradient has the unit “vertical quantity per horizontal quantity”. A volume-time graph might have litres per minute; a cost-count graph might have dollars per item. Interpreting the unit often reveals whether the gradient calculation answers a meaningful question.

6. The intercept tells us the output at zero input

In y = mx + c, setting x = 0 gives y = c. The y-intercept is therefore the point (0, c). It is a point, while c itself is the vertical coordinate of that point.

For y = 2x − 3, the y-intercept is (0, −3). To find the x-intercept, set y = 0 and solve 2x − 3 = 0, giving x = 3/2. The x-intercept is (3/2, 0).

The two calculations answer different questions. A student who substitutes x = 0 when asked for the x-intercept is confusing the name of the axis with the coordinate that vanishes there.

Worked example: for y = −2x + 3, the y-intercept is (0, 3). The x-intercept satisfies 0 = −2x + 3, so x = 1.5. The line decreases as x increases and crosses both positive axes.

In a model, the y-intercept may represent an initial quantity or a fixed charge, but only if x = 0 is meaningful and the model applies there. A line fitted to measurements for adult trees does not automatically supply a meaningful prediction at “zero years” simply because its algebra has an intercept.

Use the intercept as a mathematical feature first. Then ask whether the underlying context permits that interpretation. A representation can extend beyond the situation it was designed to describe.

7. Find a straight-line equation from information

Two distinct points determine a unique straight line. For a non-vertical line, find its gradient, then substitute a known point into y = mx + c to find c.

Using (−1, 5) and (3, −3), the gradient is −2. Substitute the first point:

5 = −2(−1) + c
5 = 2 + c
c = 3.

The equation is y = −2x + 3. Check the second point: −2(3) + 3 = −3. Both supplied points fit, and the gradient is the required one.

The point-slope form gives the same relationship:

y − y₁ = m(x − x₁).

With (−1, 5), this becomes y − 5 = −2(x + 1), hence y = −2x + 3. Brackets matter because the chosen coordinate is negative.

Another example: a line has gradient 3 and passes through (2, 11). Substitution into y = 3x + c gives 11 = 6 + c, so c = 5. The equation is y = 3x + 5.

A vertical line must be handled separately. Through (4, 1) and (4, 9), the line is x = 4. It cannot be written as y = mx + c with a finite real m. Recognising that exception is more reliable than forcing a division by zero into the gradient calculation.

8. Direct proportion is a special straight-line relationship

A direct proportionality has the form y = kx over the stated domain. Its line passes through the origin when zero is part of that domain. For non-zero x, the ratio y/x is constant.

The rule y = 4x is directly proportional. Doubling x doubles y. The rule y = 4x + 7 is linear but not directly proportional: doubling x does not generally double the output because the fixed term 7 remains only once.

Illustrative cost model: suppose a service charges a fixed $6 plus $1.20 per item. The model is C = 6 + 1.2n. At n = 5 the cost is $12; at n = 10 it is $18. The item count doubles, but the total cost does not.

The gradient, 1.2 dollars per item, describes the variable charge. The intercept, 6 dollars, describes the fixed charge. Those two components play different roles, and their graph makes the distinction visible.

For whole items, n is a non-negative integer. A continuous line can show the underlying cost rule, but only the points at permitted integer inputs represent actual purchases under that model. State that domain when interpreting a crossing or threshold.

The ratio and percentage guide develops the same distinction numerically: a constant rate does not imply that the entire output is a constant multiple of the input when a fixed starting amount is present.

9. An intersection solves two relationships at once

A point where two graphs intersect belongs to both relationships. Its coordinates therefore satisfy both equations. This gives a visual interpretation of simultaneous equations.

Consider y = 2x + 1 and y = −x + 7. At an intersection, the two expressions for y are equal:

2x + 1 = −x + 7
3x = 6
x = 2.

Substituting gives y = 5, so the intersection is (2, 5). A graph should show the lines meeting near that point, while the algebra provides its exact coordinates.

Parallel distinct lines with equal gradients but different intercepts do not meet. For y = 2x + 1 and y = 2x + 4, equating the outputs would require 1 = 4. There is no solution.

If two equations describe the same line, every point on that line is a common solution. For example, 2y = 4x + 2 simplifies to y = 2x + 1. Plotting both does not create two independent relationships; it repeats the same one.

A graphical reading has the precision of the drawing and its scale. If a crossing appears to be at x ≈ 1.7, do not report many additional decimal places that the graph cannot support. Use algebra when an exact value is required and the equations are available. The visual and symbolic methods should support each other rather than make incompatible claims about precision.

10. Graphs can solve equations with a target level

To solve f(x) = a graphically, locate the points on y = f(x) whose vertical coordinate is a. On a sketch, that corresponds to intersections with the horizontal line y = a. Read the x-values of those intersections.

For y = 2x − 3, solving 2x − 3 = 5 means finding where the line reaches y = 5. Algebra gives x = 4. The graph’s point is (4, 5), but the answer to the equation is the input 4.

To solve f(x) = g(x), compare the two graphs directly, as in the preceding section. This is not the same as setting both functions equal to zero. The graphs can intersect above or below the x-axis.

For a quadratic, a horizontal target may produce two solutions, one or none. The number of crossings depends on the target and the curve’s shape. A graph can help anticipate the number of candidates before an algebraic calculation begins.

Reading check: if a graph crosses y = 6 at x = −1 and x = 5, the solutions to f(x) = 6 are −1 and 5, subject to the stated domain. The output 6 is the target, not a third solution. If the domain is x ≥ 0, only the permitted crossing at 5 remains.

11. A quadratic is not a straight line between a few dots

Consider y = x² − 4x + 3. A short table is:

The first differences for equal input steps are −3, −1, 1 and 3, so the gradient is not constant. The graph is a curve rather than a single straight line through the points.

Two algebraic forms reveal different features:

y = (x − 1)(x − 3)
y = (x − 2)² − 1.

The factorised form gives the x-intercepts (1, 0) and (3, 0). The completed-square form shows that the minimum occurs at x = 2, where y = −1, because a real square is non-negative. The axis of symmetry is x = 2. The original form gives the y-intercept (0, 3).

These are not three different quadratics. They are three representations of the same expression, each making a feature easier to read. The algebra guide supplies the expansion and factorisation checks connecting them.

When sketching, plot the intercepts and turning point, then draw a smooth curve consistent with the equation. Do not join a sparse table using sharp straight-line segments unless the problem explicitly describes a piecewise-linear relationship. The way points are joined is a claim about what happens between them.

12. Extension: reciprocal relationships have excluded inputs

The rule y = 12/x is not defined at x = 0. For positive inputs, a useful table is:

The product xy remains 12, but the gradient is not constant. Doubling x halves y. This is an inverse proportionality rather than a direct proportionality.

If the relationship models the length and width of a rectangle with area 12 square units, both x and y must be positive. The negative branch of the bare mathematical relation is not a permitted rectangle under that model, even though negative x-values give mathematically defined negative outputs.

For non-zero real x, y never equals zero because 12/x cannot be zero. The graph approaches the axes in the relevant directions but does not include points on them. Do not join the two branches through the forbidden input zero.

A different kind of exclusion occurs in y = (x² − 9)/(x − 3). For x ≠ 3, the rule simplifies to y = x + 3. Its graph follows that straight line except that the point (3, 6) is excluded. The simplified expression does not erase the original restriction.

These examples teach one shared discipline: the graph must show the relationship’s domain, not just the shape suggested by a simplified formula. A visually smooth line is not permission to include a forbidden point.

13. A journey graph needs a precise vertical quantity

The phrases “distance travelled” and “distance from home” do not describe the same quantity. Total distance travelled cannot decrease as a journey proceeds. Distance from home can decrease while someone returns home.

Consider this invented journey along a straight route:

Assume constant movement on each travel segment and no movement during the middle segment. From 0 to 2 minutes, the distance from home increases at 120/2 = 60 metres per minute. From 2 to 5 minutes, the graph is horizontal, representing a stop. From 5 to 8 minutes, the distance from home decreases at 120/3 = 40 metres per minute.

The final segment has gradient −40 metres per minute. That is a signed rate of change of distance from home, not a negative amount of total distance travelled. The return speed is 40 metres per minute under the stated straight-route assumptions.

Total distance travelled is 120 + 120 = 240 metres. Over the full eight-minute journey, average speed is 30 metres per minute. The traveller’s net displacement is zero because the journey ends at its starting point.

A graph of accumulated distance travelled would instead rise from 0 to 120, remain at 120 during the stop, and rise to 240 during the return. The same journey has different graph shapes because the vertical quantities differ. Read the label before interpreting an increasing or decreasing segment.

14. Piecewise rules describe different behaviour on different intervals

The preceding journey can be written as a piecewise rule for distance from home d at time t:

d = 60t for 0 ≤ t ≤ 2;
d = 120 for 2 ≤ t ≤ 5;
d = 320 − 40t for 5 ≤ t ≤ 8.

At each shared endpoint, the neighbouring expressions agree. At t = 2, the first gives 120, matching the middle segment. At t = 5, the final rule gives 120, also matching. At t = 8, it gives zero.

The final expression is not simply −40t. Its negative gradient describes the rate of return, but its constant term places that segment at the correct height and time. A rate alone is insufficient to determine a line’s position.

To evaluate the model at t = 6, first select the interval containing 6. Use d = 320 − 40(6) = 80 metres. Using the outward rule 60t at every time would ignore the fact that the journey changed behaviour.

Piecewise models also appear in charges with thresholds, tanks that fill and pause, or quantities whose governing rule changes. The boundaries are part of the model. Whether an endpoint is included on one side or the other must be made clear, particularly when the rule has a jump.

Here the model is continuous at the joins, so overlapping endpoint notation causes no conflict. In a model with different values at the same boundary, define a single output for each permitted input if the relationship is intended to be a function.

15. Data points, fitted models and certainty are different things

An exact equation such as y = 3x + 2 defines a relationship. A collection of measured points may merely suggest that a similar line is useful. Measurement error, omitted variables and limited observations can all make a fitted model approximate rather than exact.

When a question stipulates an exact rule, calculate from that rule. When it supplies observations and asks for an estimate, state the estimation method and avoid reporting more precision than the data justify. A best-fit line need not pass through every observed point.

Interpolation estimates within the observed input range. Extrapolation extends beyond it. Neither automatically guarantees accuracy, but extrapolation adds the assumption that the relationship continues into a region not represented by the observations.

For example, observations of plant height over a short interval might be approximately linear. Extending that line indefinitely would imply unbounded growth at the same rate. The algebra permits the calculation; the biological situation does not thereby promise it. This is a hypothetical modelling caution, not a report of an actual growth study.

Also separate association from cause. A graph showing that two quantities vary together does not, by itself, establish that changing one will cause the other to change. The graph represents the supplied relationship; evidence for a causal explanation requires more than the plotted pattern.

A useful model report states the quantities, units, domain, rule, assumptions and limits. These are not optional disclaimers after the mathematics. They specify what the mathematics is about.

16. Capstone: compare two invented pricing rules

Suppose two hypothetical plans charge for whole items, with no tax, discount, cap or other fee in this model. Plan A costs A = 6 + 1.2n dollars. Plan B costs B = 12 + 0.8n dollars. The allowed item counts are the integers from 0 to 40. Which plan is cheaper at each count?

Plan A has the lower fixed charge but the higher per-item rate. Plan B begins higher and rises more slowly. That difference suggests a crossing, but the exact threshold should be calculated:

6 + 1.2n = 12 + 0.8n
0.4n = 6
n = 15.

At 15 items, both cost $24. For fewer than 15 items, Plan A is cheaper. For more than 15 items within the stated range, Plan B is cheaper. With whole-number counts, that means A for 0 through 14, a tie at 15, and B for 16 through 40.

A difference equation makes the result transparent:

A − B = −6 + 0.4n.

A negative difference means A costs less; a positive difference means B costs less. The sign changes at n = 15. This agrees with the graphical interpretation of the two lines crossing at (15, 24).

Check two cases. At n = 5, A = $12 and B = $16. At n = 25, A = $36 and B = $32. These checks support the interpretation on either side of the threshold. The algebraic difference establishes the full comparison over the allowed domain.

Do not extend this answer to real purchasing decisions without confirming the actual charging terms. The useful lesson is how fixed amounts, rates, intersections and domain constraints work together in a declared model.

17. Independent practice

Sketch on squared paper where helpful. Show the equation or calculation supporting a graphical answer, and distinguish exact results from values estimated from a drawing.

1. Which axis contains (0, −6), and which contains (5, 0)?
2. For y = 3x − 2, find y when x = −1, 0 and 2.
3. Find the gradient through (1, 4) and (5, 16).
4. Find the gradient through (−2, 7) and (4, −5).
5. State the gradient of y = 9 and describe the gradient of x = 3.
6. Find both intercepts of y = 2x − 8.
7. Find the equation of a line with gradient 4 through (2, 11).
8. Find the equation through (0, 5) and (3, −1).
9. Decide whether y = 7x and y = 7x + 2 are directly proportional relationships.
10. Solve the intersection of y = 3x + 1 and y = x + 9.
11. Explain why y = −2x + 3 and y = −2x − 5 do not intersect.
12. A model gives C = 4 + 2.5n. Interpret the gradient and intercept when C is dollars and n is a count of items.
13. For y = x² − 6x + 8, find its x-intercepts and turning point.
14. For y = 18/x, find y at x = 3 and state an excluded input.
15. A distance-from-home graph falls from 150 metres to zero over five minutes of a straight return journey at constant speed. Find its gradient and the return speed.
16. A graph uses two horizontal units per square and three vertical units per square. A line rises five squares over a run of three squares. Find its numerical gradient.
17. Two groups of plotted points lie on y = 2x + 1 and 2y = 4x + 2. How many common points do the full lines have?
18. Plans have costs P = 10 + 2n and Q = 22 + n. Find the crossing and identify the cheaper plan for n = 20.
19. Simplify y = (x² − 16)/(x − 4) and state which point is excluded from its straight-line representation.
20. A table contains the pairs (0, 1), (1, 3) and (2, 5). Does the table alone prove that the rule is y = 2x + 1 for every real x? Explain.

18. Worked answers

1. The y-axis and x-axis respectively. The zero x-coordinate places (0, −6) on the vertical axis. The zero y-coordinate places (5, 0) on the horizontal axis.

2. −5, −2 and 4. Substitute each complete input into 3x − 2. In particular, 3(−1) − 2 is −5, not 1. The resulting coordinate pairs are (−1, −5), (0, −2) and (2, 4).

3. Gradient 3. The vertical change is 16 − 4 = 12 and the horizontal change is 5 − 1 = 4. Their quotient is 3.

4. Gradient −2. The changes are −5 − 7 = −12 and 4 − (−2) = 6. A consistent ordering gives −12/6 = −2.

5. Zero; undefined. The first line is horizontal, with no change in y. The second is vertical, with zero horizontal change, so the usual gradient quotient is undefined.

6. (0, −8) and (4, 0). Set x = 0 for the y-intercept. Set y = 0 and solve 2x − 8 = 0 for the x-intercept. State each answer as a point.

7. y = 4x + 3. Substitute (2, 11) into y = 4x + c: 11 = 8 + c. The constant is 3, not 11.

8. y = −2x + 5. The gradient is (−1 − 5)/(3 − 0) = −2. The supplied point with x = 0 directly gives the intercept 5.

9. Only y = 7x is directly proportional. The other rule has a non-zero fixed term. For example, doubling x from 1 to 2 changes its output from 9 to 16, not to 18.

10. (4, 13). Equate the outputs: 3x + 1 = x + 9 gives x = 4. Substitute into either original equation to obtain y = 13, then verify the other.

11. They are distinct parallel lines. Equal gradients and different intercepts give no common point. Algebraically, equality would require 3 = −5.

12. A variable charge of $2.50 per item and a fixed charge of $4. These meanings belong to the stated model. The model’s actual purchase inputs are whole-number item counts.

13. Intercepts (2, 0) and (4, 0); turning point (3, −1). Factorise to (x − 2)(x − 4) for the roots. Completing the square gives (x − 3)² − 1 for the minimum.

14. y = 6; x = 0 is excluded. Divide 18 by 3. The expression has no defined real quotient at a zero denominator. A positive-length context would impose further positivity restrictions.

15. Gradient −30 m/min; speed 30 m/min. Distance from home decreases by 150 metres over five minutes. Under the given straight-route assumptions, the speed is the magnitude of that rate.

16. Gradient 2.5. The rise is 5 × 3 = 15 vertical units, and the run is 3 × 2 = 6 horizontal units. Divide 15 by 6. Counting only squares would ignore the axes’ scales.

17. Infinitely many. Dividing the second equation by 2 gives the first. They describe the same full line, not two independent lines with just one crossing.

18. Crossing at (12, 34); Q is cheaper at n = 20. Equating gives 10 + 2n = 22 + n, so n = 12. At twenty items, P = 50 and Q = 42.

19. y = x + 4 for x ≠ 4; the point (4, 8) is excluded. Factorisation cancels x − 4 only where that factor is non-zero. The simplified formula does not restore the forbidden input.

20. No. The line fits the three pairs, but y = 2x + 1 + x(x − 1)(x − 2) fits them too and differs at other inputs. Additional information—such as a stated linear rule—is needed for the universal conclusion.

19. Diagnose a graph mistake by its first cause

A wrong graph may begin with incorrect substitution, reversed coordinates, an irregular scale, a lost unit, an unsuitable joining rule or a misread domain. These problems require different repairs. Redrawing the entire graph several times without locating the first cause can leave the misunderstanding intact.

Check the table against the equation before plotting. Check the axis labels and scale before positioning points. Check special features such as intercepts or symmetry against the algebra. Finally, read one point back into the original context in a complete sentence.

For an intersection question, verify both relationships. For a journey question, name the vertical quantity and distinguish signed gradient from speed. For a fitted model, distinguish observations from the proposed continuation. For a formula with a denominator, inspect excluded inputs before drawing through them.

A suggested practice sequence is one table-to-graph task, one graph-to-sentence interpretation, and one equation-to-feature task. Later, reverse the direction: provide a gradient and a point, or a context with a fixed amount and a rate, and ask the learner to produce the equation. This is a teaching suggestion rather than a measured guarantee of learning.

The aim is not simply a tidy graph. It is a representation whose points, scales, conditions and interpretation agree with the relationship being studied.

20. Continue through the BTT Mathematics library

Return to the BTT Mathematics Hub for the wider collection, or use the BTT Mathematical Lab to investigate recurring representation errors. The broader Secondary 1 Mathematics guide provides the transition doorway rather than replacing this worked graph companion.

Within this batch, use Signed Numbers, Brackets and Algebraic Structure for substitution and equivalent forms, Equations, Balance and Checking for intersections and exact solutions, and Ratio, Percentage and the Correct Base for rates, proportionality and meaningful comparisons.

Sources and scope

The displayed mathematical rules, tables, derivations, counterexamples and practice questions are original teaching material. The example journeys and charging plans are hypothetical, with their assumptions stated. No model is presented as an empirically verified description of a real service or journey.

Consult MOE: Curriculum for secondary schools for the relevant school and subject-level framework. This guide’s mathematical progression is not an assertion that every extension is required in every secondary year or subject level. Framework checked on 6 September 2026.