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How do we model a situation whose rate of change shifts at certain points?

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

How to build a piecewise-linear model where different straight-line rules apply over different intervals, graph it, read its breakpoints, and use it to solve practical step-rate problems.

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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).

Choosing the right piece

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

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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20234 marksCalculator-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.

SACE 20223 marksCalculator-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.

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