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Piecewise-linear models (SACE Stage 2 General Mathematics)

Syllabus dot point

“Construct, graph and interpret piecewise-linear models in which the rule changes over different intervals of the domain.”

8 min read

Quick answer

A piecewise-linear model joins two or more straight-line segments, each with its own rule, that apply over different intervals of xx. You choose the correct segment for the input you are given, then substitute into that segment's equation.

Jump to a section
  1. What this dot point is asking
  2. What a piecewise-linear model is
  3. Choosing the right piece
  4. Graphing the model
  5. Reading information back from the model
  6. Why the second piece carries a constant
  7. Solving equations with a piecewise model
  8. Exam-style questions

What this dot point is asking

You must be able to write a model in pieces, state the interval each piece applies to, graph the segments, identify the breakpoints where the rule changes, and use the model to calculate and interpret values.

What a piecewise-linear model is

Many real charges and rates do not follow a single straight line. A taxable income, a phone plan, or a delivery fee may follow one rate up to a threshold, then a different rate beyond it. We write this as several linear rules, each tagged with the interval where it applies:

C(x)={2x+10,0≤x≤203x−10,x>20 C(x) = \begin{cases} 2x + 10, & 0 \le x \le 20 \\ 3x - 10, & x > 20 \end{cases}

Each piece is just y=mx+cy = mx + c. The values of xx where the rule changes are the breakpoints (here x=20x = 20).

Breakpoint

A breakpoint is an xx-value where the model switches from one linear rule to the next. A model is continuous if the two pieces give the same yy-value at the breakpoint, so the graph has no jump.

Choosing the right piece

The single most important skill is matching the input to its interval before you substitute.

Worked example

A two-rate electricity charge

An electricity retailer charges 0.30 dollars per kWh for the first 300 kWh in a quarter, then 0.45 dollars per kWh for any usage above 300 kWh. Find the cost of using 500 kWh.

The first 300 kWh cost a flat block. Build the model for total cost CC against usage uu:

C(u)={0.30u,0≤u≤3000.45(u−300)+90,u>300 C(u) = \begin{cases} 0.30u, & 0 \le u \le 300 \\ 0.45(u - 300) + 90, & u > 300 \end{cases}

The constant 9090 is the cost of the first 300 kWh, since 0.30×300=900.30 \times 300 = 90. For u=500u = 500 we use the second piece because 500>300500 > 300:

C(500)=0.45(500−300)+90=0.45(200)+90=90+90=180C(500) = 0.45(500 - 300) + 90 = 0.45(200) + 90 = 90 + 90 = 180

The cost is 180 dollars.

Graphing the model

Plot each segment only over its own interval. At a breakpoint, draw a filled dot for the endpoint that is included and an open dot for one that is excluded. A change in gradient appears as a corner ("kink") in the graph; a steeper gradient after the breakpoint means the quantity grows faster beyond that point.

Continuity at a breakpoint

A piecewise model is continuous if both pieces give the same value at the breakpoint. For the electricity model, the first piece gives 0.30(300)=900.30(300) = 90 and the second gives 0.45(0)+90=900.45(0) + 90 = 90, so the graph joins with no gap.

Reading information back from the model

Examiners often ask you to interpret the graph in words:

  • Each segment's gradient is the rate over that interval (cost per kWh, fee per kilometre).
  • A breakpoint marks where conditions change (a usage threshold, a tax bracket boundary).
  • A horizontal segment (m=0m = 0) means the quantity stays constant over that interval, such as a fixed monthly fee regardless of usage.
Worked example

Reading a delivery-fee graph

A courier graph shows a flat fee of 15 dollars for distances from 0 to 10 km, then a line rising from 15 dollars at 10 km to 35 dollars at 30 km. Find the per-km rate beyond 10 km.

Beyond the breakpoint the cost rises from 15 to 35 dollars as distance goes from 10 km to 30 km:

m=35−1530−10=2020=1.m = \frac{35 - 15}{30 - 10} = \frac{20}{20} = 1.

So beyond 10 km the fee increases by 1 dollar per kilometre.

Why the second piece carries a constant

In the electricity model the second rule is 0.45(u−300)+900.45(u - 300) + 90, not simply 0.45u0.45u. The (u−300)(u - 300) measures usage beyond the threshold, and the +90+90 is the accumulated cost of the first 300 kWh. This structure - a base amount plus a new rate applied only to the excess - is exactly how tax brackets and tiered tariffs work in practice. Writing the excess term explicitly keeps the model continuous and stops you from wrongly charging the higher rate on the whole amount. It is the single most important idea for piecewise charge models.

Solving equations with a piecewise model

To find when C(x)C(x) equals a target value, decide which interval the answer falls in, then solve that piece's equation. Always check the solution actually lies in that interval; if it does not, the answer belongs to a different piece.

State the interval with every answer

When you give a piecewise value, name the interval you used. Writing "since 500>300500 > 300 I use the second rule" shows the examiner your reasoning and earns method marks even if your arithmetic slips.

Common mistake

Do not substitute an input into the wrong piece. A frequent error is putting u=500u = 500 into the first rule 0.30u0.30u, giving 150150 instead of 180180. Always check which interval the input belongs to before substituting, and never apply the high rate to the entire amount when only the portion above the threshold is charged at that rate.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Calculator-assumed. An electricity retailer charges 0.30 dollars per kWh for the first 300 kWh in a quarter, then 0.45 dollars per kWh for usage above 300 kWh. (a) Write a piecewise model for the total cost CC dollars against usage uu kWh. (b) Find the cost of using 500 kWh.
Show worked answer →

(a) The first 300 kWh cost 0.30×300=900.30 \times 300 = 90 dollars, so the model is

C(u)={0.30u,0≤u≤3000.45(u−300)+90,u>300.C(u) = \begin{cases} 0.30u, & 0 \le u \le 300 \\ 0.45(u - 300) + 90, & u > 300. \end{cases}
(2 marks)

(b) Since 500>300500 > 300, use the second piece: C(500)=0.45(200)+90=90+90=180C(500) = 0.45(200) + 90 = 90 + 90 = 180 dollars. (2 marks)

Marks: two for a correct two-piece model with the 90 carried forward, two for selecting the right piece and computing 180.

Practice3 marks
Calculator-assumed. A courier charges a flat 15 dollars for distances from 0 to 10 km, then a rate per km beyond 10 km so that a 30 km trip costs 35 dollars. (a) Find the per-km rate beyond 10 km. (b) Determine the cost of a 22 km trip.
Show worked answer →

(a) Beyond 10 km the cost rises from 15 to 35 dollars as distance goes 10 km to 30 km: m=35−1530−10=2020=1m = \dfrac{35 - 15}{30 - 10} = \dfrac{20}{20} = 1 dollar per km. (1 mark)

(b) For d=22d = 22 (beyond the breakpoint), cost =15+1(22−10)=15+12=27= 15 + 1(22 - 10) = 15 + 12 = 27 dollars. (2 marks)

Marks: one for the gradient beyond 10 km, two for applying the correct piece to find 27.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

ExamExplained