Mathematics • Year 9 • Unit 2 • Lesson 1

Linear or Curved? Real-World Decisions

Decide which family fits each situation: a ball falling under gravity, a bedroom shared between siblings, money in a savings account, area of a square garden, and a phone screen budget. Then explain your reasoning in plain Year 9 English.

Apply · Real-World Maths

1. Word problems

For each scenario: build a table if asked, decide linear or non-linear, name the family if non-linear, and show your evidence (first differences, the equation form, or both). 3 marks each

1.1 — Falling ball. A ball is dropped from a balcony. Its distance fallen $d$ (m) after $t$ seconds is recorded: $t = 0, 1, 2, 3$ gives $d = 0, 5, 20, 45$.

(a) Compute the first differences in $d$.
(b) State whether the relationship is linear or non-linear.
(c) Explain in one sentence why this matches what you know about gravity making objects accelerate.

Stuck? If the first differences were all the same number, the ball would be falling at a constant speed. It isn't.

1.2 — Pizza for the table. One large pizza is shared equally between $n$ people at the table. Each person gets $s = 12/n$ slices.

(a) Build a table of $s$ for $n = 1, 2, 3, 4, 6, 12$.
(b) State which family of curve $s = 12/n$ belongs to.
(c) Describe in one sentence what happens to the slices per person as more people join.

Stuck? $s = 12/n$ has $n$ in the denominator — that's the signature of one of the three families.

1.3 — Square garden area. The side length of a square garden is $L$ metres. Its area $A$ in m² is given by $A = L^2$.

(a) Complete the table for $L = 1, 2, 3, 4, 5$ metres.
(b) Show that the first differences in $A$ are not constant.
(c) Name the family of curve and state which way it opens.

Stuck? Doubling the side does not double the area — it quadruples it. That's the parabola signature in disguise.

1.4 — Birthday savings. Tom adds $\$25$ to his savings account each month, starting with $\$100$. After $m$ months his balance is $B = 100 + 25m$.

(a) Compute his balance after $m = 0, 1, 2, 3, 4$ months.
(b) Compute first differences in $B$.
(c) State whether the relationship is linear or non-linear and give the reason.

Stuck? Equal monthly deposits = equal jumps in the balance = constant first differences = one of the families.

1.5 — Doubling rumour. One student starts a rumour at 8 a.m. Every hour the number of students who know it doubles. Let $h$ be the number of hours since 8 a.m., so the number who know is $N = 2^h$.

(a) Compute $N$ for $h = 0, 1, 2, 3, 4, 5$.
(b) Show that the first differences in $N$ are NOT constant.
(c) Name the family of curve.

Stuck? "$x$ is in the exponent" $\Rightarrow$ one specific family from the lesson.

2. Explain your thinking

Use full sentences, no dot points. 4 marks

2.1 A classmate looks at the equation $y = 5x + 7$ and says "it has an $x$ in it, so it must be non-linear." In your own words, explain (i) why their reasoning is wrong, (ii) which feature of the equation actually decides whether a relationship is linear, and (iii) give one example of a non-linear equation that they could have looked at instead.

Stuck? Revisit lesson § "Spot the Trap" — the first row is exactly this misconception.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Falling ball

(a) First differences in $d$: $5 - 0 = 5$, $20 - 5 = 15$, $45 - 20 = 25$. So $5, 15, 25$.
(b) Non-linear — the first differences are not constant; in fact they grow by $10$ each second (constant second differences indicate a parabola). The rule is $d = 5t^2$.
(c) Gravity makes the ball accelerate, so the distance fallen in each successive second gets larger — matching a curved (parabolic) graph rather than the straight line we'd see for constant speed.

1.2 — Pizza for the table

(a) $n = 1, 2, 3, 4, 6, 12$ gives $s = 12, 6, 4, 3, 2, 1$.
(b) $s = 12/n$ is a $k/x$ pattern, so the curve family is the hyperbola.
(c) As more people join, the slices per person decrease quickly at first, then more slowly — doubling the people from $1$ to $2$ cuts the slices in half (12 to 6), but going from $6$ to $12$ only cuts from $2$ to $1$.

1.3 — Square garden area

(a) $L = 1, 2, 3, 4, 5$ gives $A = 1, 4, 9, 16, 25$ m².
(b) First differences: $3, 5, 7, 9$ — not constant.
(c) Curve family: parabola (the equation $A = L^2$ has $L^2$). Because the coefficient of $L^2$ is positive ($+1$), the parabola opens upward.

1.4 — Birthday savings

(a) $m = 0$: $\$100$. $m = 1$: $\$125$. $m = 2$: $\$150$. $m = 3$: $\$175$. $m = 4$: $\$200$.
(b) First differences: $25, 25, 25, 25$.
(c) Linear — the first differences are constant ($25$ every month), which is exactly the straight-line signature. The equation $B = 100 + 25m$ is already in $y = c + mx$ form.

1.5 — Doubling rumour

(a) $N = 1, 2, 4, 8, 16, 32$.
(b) First differences: $1, 2, 4, 8, 16$ — not constant (each one is double the last).
(c) Curve family: exponential — $N = 2^h$ has $h$ as the exponent.

2.1 — Explain your thinking (sample response)

My classmate is wrong because "having an $x$" is not what makes a relationship non-linear — every linear equation has an $x$ in it too. What actually decides is what is done to $x$. For an equation to be linear, $x$ must appear only to the power $1$, not squared, not in a denominator, and not as an exponent. In $y = 5x + 7$ the $x$ is to the power $1$ and is just multiplied by $5$, so the graph is a straight line with gradient $5$ and $y$-intercept $7$. A non-linear equation they could have looked at instead is $y = x^2$, where the squared $x$ produces a parabola.

Marking: 1 mark for "linear equations also have $x$"; 1 mark for "$x$ to the power 1 only"; 1 mark for a valid non-linear example; 1 mark for clear, full-sentence writing.