Lesson 29 · The Stable Story · Chapter 4

The Grant Committee Wants Proof

Table. Graph. Equation. One convincing account of what the stable needs.

26 · Audit27 · Build28 · Investigate29 · Present30 · Organise31 · Open
Two days after the sensor alarm…

The inspection finds a faulty valve. The repair and a better monitor will cost more than the €3,321 remaining from the build, so Mia asks the town animal-welfare fund for help. The committee will not accept “the graph looked strange.” It wants evidence that can be checked in three forms.

Mia's task is to make each table, graph, and equation agree. If one view contradicts another, the proposal loses credibility; if all three match, the committee can see both the pattern and its meaning.

Mia presents the stable's table, graph, and equation evidence to the community grant committee.

A table, graph, and equation are not three separate answers. They are three windows looking at the same relationship. A strong solution explains what the shared rate and starting value mean for the stable.

The translation bridge

Table
$x$012
$y$4710
see repeated changes
Graph
↗
see a straight line starting at $4$
Equation

$y=3x+4$

rate $3$, start $4$

The table adds $3$ each step. The graph rises $3$ for every run of $1$. The equation places $3$ beside $x$. In every view, the rate is $3$.

Worked example · Water for the horses

A tank starts with $30$ L and loses $2$ L each hour.

Write the equation. Start at $30$ and change by $-2$: $y=-2x+30$.
Build the table. For $x=0,3,6$, the values are $30,24,18$.
Picture the graph. Plot $(0,30)$, $(3,24)$, and $(6,18)$. The straight line falls because the signed rate is negative.
Interpret. Water decreases by $2$ litres per hour.

Practice · Translate carefully

Translate in every direction
1 · Tack-room shelves. A table gives $(x,y)=(0,3),(2,7),(4,11),(6,15)$, where $x$ is shelves built and $y$ is hours used.
(a) Calculate the shared rate for the intervals.
(b) Write the equation.
(c) State the graph's starting point.
(d) Choose whether the graph rises or falls.
table changes → rate → x = 0 value → equation → graph shape
2 · Valve-repair sealant. A straight graph contains $(0,30)$, $(3,24)$, and $(6,18)$, where $x$ is joints sealed and $y$ is sealant remaining in millilitres.
(a) Calculate the signed rate.
(b) Write the equation.
(c) Interpret the rate with units and explain what it predicts for the repair team.
graph direction → negative rate → intercept → equation → material-use meaning
3 · Market delivery. The equation is $y=1.5x+4$, where $x$ is kilometres travelled and $y$ is delivery cost in euros.
(a) Calculate $y$ when $x=0$.
(b) Calculate $y$ when $x=2$.
(c) Calculate $y$ when $x=6$.
(d) State the rate with units.
(e) State the starting value with units.
(f) Name the three coordinates to plot.
substitute → ordered pairs → read m and b → predict the graph
4 · Do not promise an unreliable donation rate. A social-media fundraiser records $(0,2),(1,5),(2,10),(3,17)$, where $x$ is days and $y$ is total donations in tens of euros.
(a) Calculate the three interval rates.
(b) Decide LINEAR or NOT LINEAR.
(c) Explain why Mia must not use one equation of the form $y=mx+b$ to promise the committee a fixed daily donation rate.
Meaning: a model is useful only when its assumption—one constant rate—is supported by the evidence.
test every interval → find changing rates → reject one constant m → protect the proposal
From here, choose the most useful view yourself.
5 · Compare two monitoring services. Service A's table contains $(0,12),(2,20),(4,28)$. Service B follows $y=5x+7$. In both, $x$ is months and $y$ is total cost in euros.
(a) Calculate Service A's monthly rate.
(b) Write Service A's equation.
(c) State Service B's monthly rate.
(d) State Service B's starting cost.
(e) Calculate Service A's cost for $6$ months.
(f) Calculate Service B's cost for $6$ months.
(g) Choose the cheaper service for the six-month grant request and state the saving.
Decision: this comparison becomes a real line in the committee budget.
translate both forms → compare at one shared input → decide
6 · Add one honest prediction to the proposal. Invent a stable situation with a starting value of $10$ and a constant rate of $-2$. It could describe water, feed, battery charge, or another quantity that genuinely decreases.
(a) Name $x$ and $y$, including units.
(b) Write its equation.
(c) Create a coordinate pair using $x=0$.
(d) Create a coordinate pair using $x=2$.
(e) Create a coordinate pair using $x=4$.
(f) Explain how the negative rate would look on its graph and what it means in the stable.
choose a sensible decrease → equation → points → graph description → operational meaning

Spaced review

7 · Protect the repair reserve. Mia deposits €800 of stable donations at $3\%$ simple interest for $4$ years. (a) Calculate the interest. (b) Calculate the final repair reserve.
percent → decimal → I = Prt → add principal
8 · Compress the sensor archive. A sensor-image file is compressed to half its previous size each round: $320,160,80,\ldots$ KB. (a) State $r$. (b) Write an explicit formula. (c) Calculate $a_6$ with units.
ratio → formula with n − 1 → substitute
9 · Paddock check. A rectangular paddock is $18$ m by $11$ m. (a) Calculate its perimeter. (b) Calculate its area, with the correct units.
boundary uses metres → surface uses square metres
10 · Check the reserve tank. A graph passes through $(2,48)$ and $(6,36)$, where $x$ is days and $y$ is reserve-water depth in centimetres. (a) Calculate $\Delta y$. (b) Calculate $\Delta x$. (c) Calculate and interpret the rate.
ending minus starting → preserve sign → cm/day → direction
Show answers
Problem 1a
Each rate is $\frac{4}{2}=2$ hours/shelf.
Problem 1b
$y=2x+3$.
Problem 1c
It starts at $(0,3)$.
Problem 1d
The graph rises.
Problem 2a
$m=\frac{18-30}{6-0}=\frac{-12}{6}=-2$ mL/joint.
Problem 2b
$y=-2x+30$.
Problem 2c
Sealant remaining decreases by $2$ millilitres per joint, so the team can predict how much material a planned number of joints will consume.
Problem 3a
$y(0)=4$.
Problem 3b
$y(2)=7$.
Problem 3c
$y(6)=13$.
Problem 3d
Rate: €1.50/km.
Problem 3e
Starting value: €4.
Problem 3f · point 1
$(0,4)$.
Problem 3f · point 2
$(2,7)$.
Problem 3f · point 3
$(6,13)$.
Problem 4a · interval 1
$3$ tens of euros/day.
Problem 4a · interval 2
$5$ tens of euros/day.
Problem 4a · interval 3
$7$ tens of euros/day.
Problem 4b
NOT LINEAR.
Problem 4c
A linear equation needs one constant $m$, but these interval rates change.
Problem 5a
Service A: $m=\frac{20-12}{2-0}=4$ €/month.
Problem 5b
Service A: $y=4x+12$.
Problem 5c
Service B: €5/month.
Problem 5d
Service B's starting cost is €7.
Problem 5e
Service A: $4(6)+12=€36$.
Problem 5f
Service B: $5(6)+7=€37$.
Problem 5g
Choose Service A; it is cheaper by €1 over six months.
Problem 6a · x
Answers vary. Example: $x=$ hours.
Problem 6a · y
Example: $y=$ litres of water remaining.
Problem 6b
Example: $y=-2x+10$.
Problem 6c
Example: $(0,10)$.
Problem 6d
Example: $(2,6)$.
Problem 6e
Example: $(4,2)$.
Problem 6f
The line falls from left to right because $y$ decreases as $x$ increases.
Problem 7a
$I=800\times0.03\times4=€96$.
Problem 7b
$A=800+96=€896$.
Problem 8a
$r=\frac12$.
Problem 8b
$a_n=320\left(\frac12\right)^{n-1}$.
Problem 8c
$a_6=320\left(\frac12\right)^5=10$ KB.
Problem 9a
$P=2(18+11)=58$ m.
Problem 9b
$A=18\times11=198$ m².
Problem 10a
$\Delta y=36-48=-12$ cm.
Problem 10b
$\Delta x=6-2=4$ days.
Problem 10c
$m=\frac{-12}{4}=-3$ cm/day; the reserve-water depth decreases by $3$ cm each day.
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