050i - Problem Solving with Linear Functions

Lesson 50 — Problem Solving and Consolidation: Linear Functions and Relations

Strand: Algebra | Descriptor: AC9M8A04 | Duration: 45 minutes

Learning Intentions

• To apply understanding of linear function families to solve real-world problems.
• To consolidate skills in graphing, conjecturing and generalising linear relationships.

Success Criteria

I can:

1. Model a real-world context with a linear function of the form .
2. Compare linear functions representing different scenarios and interpret their point of intersection.
3. Test and justify a conjecture about a family of functions using evidence.
4. Communicate a generalisation clearly, using both algebra and words.

Warmup

(6 minutes — matching, pairs)

Match each context description to the feature of a linear function it corresponds to.

1. “A 50$ sign-up bonus before anything is earned.” — (gradient / intercept)
2. “Earns 12$ for every hour worked.” — (gradient / intercept)
3. “Starts at zero and grows steadily.” — (gradient / intercept)
4. “A membership that never changes in cost, however much you use it.” — (gradient / intercept)

Answers: 1. Intercept; 2. Gradient; 3. Intercept ; 4. Gradient (a horizontal, constant function).

Activities

Activity 1 — Applied Problems: Modelling with Linear Functions (12 min)

Pairs. For each context: define variables, write the function, and state what and mean physically.

Problem 1. A plumber charges a 60$45$ per hour.

Problem 2. A tank starts with L of water and drains at L per minute.

Problem 3. A phone plan has no flat fee and charges 0.08$ per MB of data.

For each, write , identify and in context, and calculate the cost/amount after a given input (e.g. hours, minutes, MB respectively).

Answers: Problem 1 — ; at , 195V = -20t + 500t=8V=340C = 0.08dd=500C=$40$.

Activity 2 — Rich Applied Investigation: Comparing Plans (14 min)

Pairs, using a digital graphing tool where available. Full Polya cycle required.

Two ride-share companies charge as follows:

• RideCo: 3$1.80$ per km.
• GoCar: 1.20$ per km, no flat fee.

A conjecture is proposed: “RideCo is always more expensive than GoCar, because it has a flat fee that GoCar doesn’t.”

1. Write a cost function for each company.
2. Test the conjecture using at least three distances, including a very short trip and a long trip.
3. Graph both functions (by hand or with a digital tool) and find the break-even distance.
4. Decide whether the conjecture is true, false, or needs refining. Justify with evidence.

Socratic scaffolding:

Prompt	Purpose
Understand the problem — what is being compared?	Total cost of each plan as a function of distance travelled.
Devise a plan — how do you test a conjecture about “always”?	Try several distances, including extreme (very small and very large) cases, not just one.
Carry out the plan — test km.	RideCo: 4.80$1.20$. RideCo is more expensive here.
Carry out the plan — test km.	RideCo: 21$12$. Still more expensive — but is the gap growing or shrinking?
Carry out the plan — find the break-even point algebraically.	. A negative distance is not physically meaningful.
Look back — what does a negative or “impossible” break-even point tell you?	The lines never cross for any realistic — RideCo’s line starts higher (bigger intercept) and rises faster (bigger gradient), so it is more expensive for every real distance.
Look back — is the original conjecture true, false, or refined?	True, but for a more precise reason than “it has a flat fee” — it is because RideCo’s gradient is also larger, not the flat fee alone.

Answers: RideCo ; GoCar . RideCo costs more at every tested distance; algebraically, the lines never cross for , confirming RideCo is always more expensive. The conjecture is essentially true, but the reason needs refining: it isn’t only the flat fee, but also the steeper gradient, that guarantees this.

Activity 3 — Quick Generalisation Share (7 min)

Whole class, rapid-fire.

Each pair states one generalisation from today’s or the previous three lessons’ work (e.g. about parallel lines, shared intercepts, or how gradient/intercept affect real contexts). Teacher records a master list of class generalisations about linear functions on the board as a consolidated summary of the AC9M8A04 unit.

Checks for Understanding

(6 minutes — exit ticket, collected)

1. A courier charges 8$2.506$ kg parcel.
2. Two functions are and . Find where they intersect, and state which is greater for .
3. A conjecture claims “a function with a bigger gradient is always more expensive at every input value.” Give a counterexample using two functions of your choice.
4. Reasoning. Explain, using the RideCo/GoCar investigation, why testing only one value of would have been insufficient to properly test the conjecture.

Answers: 1. ; at , 235x+2=3x+10 \Rightarrow x=4, y=22x>4y=5x+2y=5x+1y=2x+20x=1622x$; 4. A single test value cannot reveal whether a relationship (like “always more expensive”) holds across the entire range of realistic inputs — the gap between two linear functions can change or even reverse if the lines cross, which only shows up by testing multiple points or solving algebraically for the intersection.

Common Misconceptions

Misconception	How to pre-empt it
Believing a bigger gradient always means a bigger total, ignoring the intercept and the range of being considered.	Use CFU Q3’s counterexample explicitly: compare at small and large .
Treating a negative or non-physical break-even solution as “no answer” rather than interpreting it in context.	Model the RideCo/GoCar look-back step: a negative distance means the lines don’t cross for realistic values.
Testing a real-world conjecture with only one convenient value instead of a spread, including extremes.	Require at least three test values, including small and large, before accepting a conjecture.
Confusing “the flat fee makes it more expensive” with the complete, precise reason involving both gradient and intercept.	Explicitly separate the two parameters’ contributions in the look-back discussion.
Forgetting to state generalisations in terms of both algebra and words.	Model both forms side by side throughout Activity 3’s share-out.

Enrichment — Competition-Style Problems

E1 (Kangaroo style). Two functions are and . At what value of are their outputs equal, and what is that common value?

Answer

, .

E2 (AMC Junior style). A conjecture states: “For the family , every member has a positive -value when .” Is this always true? Justify.

Answer

Yes — at , regardless of , and is always positive. The conjecture is true for every real .

E3 (Challenge). Two companies’ cost functions are and . Determine, with full justification, for what range of each company is cheaper.

Answer

. For , test : , — Company 2 cheaper. For , test : , — Company 1 cheaper. So Company 2 is cheaper for , Company 1 for , equal at .

E4 (Challenge). A family of functions is . Find the fixed point shared by every member, and use it to explain why, for very large , the sign of alone determines whether is large and positive or large and negative.

Answer

At : , so every line passes through . For very large , the term dominates the fixed constant contribution, so if , , and if , , regardless of the shared point.

Homework

1. A landscaper charges a 50$354.5$ hours.
2. Two functions are and . Find their intersection point and state which function gives the greater value for .
3. A conjecture claims “the function with the smaller intercept is always cheaper.” Test this using and at and , and decide whether the conjecture holds.
4. Two library membership plans are (flat annual fee, unlimited borrowing) and (no fee, 0.50$ per book borrowed). Find the break-even number of books, and state which plan is cheaper below and above this number.
5. Reasoning. Explain why comparing two linear functions requires checking both the gradient and the intercept, not just one or the other, using an example from this lesson.
6. Challenge. A family of functions is . Find the shared fixed point of the family, and determine the value of for which the line also passes through .

Answers: Q1 — ; at , 207.503x+5=6x-4 \Rightarrow 3x=9 \Rightarrow x=3, y=14x<3y=3x+5x=05>-4x=2y=6(2)+1=13y=2+50=52x=20y=121y=7015=0.5b \Rightarrow b=3030F_230F_1xx=3y=3m+15-3m=15(3,15)(0,24)24 = m(0)+15-3m \Rightarrow 24=15-3m \Rightarrow m=-3$.