Mathematics • Year 9 • Unit 2 • Lesson 2
The Basic Parabola in the Real World
Use $y = x^2$ to model square areas, tile patterns, square paddocks and pendulum tables. Then explain why squaring a negative still gives a positive in everyday terms.
1. Word problems
Each problem uses the basic parabola $y = x^2$ to describe something real. Show your working, including brackets where you substitute negatives. 3 marks each
1.1 — Square garden bed. A square garden bed has side length $L$ metres and area $A = L^2$ square metres.
(a) Complete a table of $A$ for $L = 1, 2, 3, 4, 5$ m.
(b) Sketch the points on a set of axes (axes labelled, vertex marked).
(c) State why we only use $L \ge 0$ in this context, even though $y = x^2$ is defined for negative $x$.
1.2 — Tile pattern. A school is laying square tile patterns. Pattern $n$ uses $n^2$ tiles (so pattern 1 uses $1$ tile, pattern 2 uses $4$ tiles, etc.).
(a) How many tiles are in pattern $7$? Pattern $10$?
(b) Which pattern number uses exactly $64$ tiles?
(c) Why does the number of tiles grow quickly even though the side length grows only by $1$?
1.3 — Square paddock fencing. A square paddock has side length $L$ metres and diagonal walking-path length $d$ metres given by $d^2 = L^2 + L^2 = 2L^2$ (Pythagoras' theorem).
(a) Show that the formula simplifies to $d = L\sqrt{2}$.
(b) For $L = 5$ m, compute $d$ to one decimal place.
(c) Build a table of $d^2$ values for $L = 1, 2, 3, 4, 5$ — what kind of curve do the $(L, d^2)$ points form?
1.4 — Phone screen pixel grid. Sam's phone screen has square pixel blocks. A square block of side $s$ pixels contains $s^2$ pixels.
(a) How many pixels are in a $20 \times 20$ block?
(b) Plot the points $(s, s^2)$ for $s = 0, 1, 2, 3, 4, 5$.
(c) Use symmetry (or the equation) to state why the curve $y = x^2$ would give the same value at $s = -3$ as at $s = 3$, and explain why negative pixel sizes don't make physical sense.
1.5 — Pendulum period table. The period $T$ (seconds) of a small pendulum is approximately $T = \sqrt{L}$ for length $L$ metres. So $T^2 = L$ — meaning if you plot $L$ against $T$, the points form half of a parabola.
(a) Build a table: for $T = 0, 1, 2, 3$ seconds, compute $L = T^2$.
(b) Sketch the points $(T, L)$ on axes with $T$ horizontal and $L$ vertical.
(c) State the vertex of this curve and explain why we only plot $T \ge 0$ here.
2. Explain your thinking
Use full sentences. 4 marks
2.1 A friend types $-3^2$ into their calculator and gets $-9$. They expect $+9$ and think the calculator is broken. In your own words, explain (i) why the calculator's answer is actually correct given what it sees, (ii) what the friend needed to type to get $+9$, and (iii) the rule from Lesson 2 about brackets when substituting negative numbers into $y = x^2$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Square garden bed
(a) $L = 1, 2, 3, 4, 5$ gives $A = 1, 4, 9, 16, 25$ m².
(b) Plot $(1,1), (2,4), (3,9), (4,16), (5,25)$ — rising right half of a parabola, vertex $(0, 0)$.
(c) A garden bed can't have a negative side length, so although $y = x^2$ is defined for negative $x$, only $L \ge 0$ makes physical sense here. Real-world domain restricts the maths.
1.2 — Tile pattern
(a) Pattern $7$: $7^2 = \mathbf{49}$ tiles. Pattern $10$: $10^2 = \mathbf{100}$ tiles.
(b) $n^2 = 64 \Rightarrow n = \sqrt{64} = \mathbf{8}$ (positive root, since pattern numbers are positive).
(c) Each new pattern adds an entire new row AND column of tiles. Going from pattern $7$ ($49$ tiles) to $8$ ($64$ tiles) adds $15$ tiles, not just $1$. The total grows as a parabola because area scales with the square of the side.
1.3 — Square paddock
(a) $d^2 = 2L^2 \Rightarrow d = \sqrt{2L^2} = L\sqrt{2}$ (positive root, since $d$ is a length).
(b) $L = 5$: $d = 5\sqrt{2} \approx \mathbf{7.1}$ m.
(c) Table: $L = 1, 2, 3, 4, 5$ gives $d^2 = 2, 8, 18, 32, 50$. The $(L, d^2)$ points form a parabola (specifically $y = 2L^2$ — a vertical stretch of $y = x^2$).
1.4 — Phone screen pixel grid
(a) $20 \times 20 = 20^2 = \mathbf{400}$ pixels.
(b) Points $(0,0), (1,1), (2,4), (3,9), (4,16), (5,25)$ — the right half of the basic parabola.
(c) $(-3)^2 = 9 = 3^2$ because squaring kills the sign. So the maths predicts the same $y$-value at $\pm 3$ (the parabola is symmetric across the $y$-axis). But you can't have a pixel block with side $-3$ pixels — pixel counts are positive whole numbers, so we restrict to $s \ge 0$ in this context.
1.5 — Pendulum period table
(a) $T = 0, 1, 2, 3$ gives $L = 0, 1, 4, 9$ m.
(b) Points $(0,0), (1,1), (2,4), (3,9)$ — right half of a parabola, vertex at origin.
(c) Vertex $(0, 0)$. We only plot $T \ge 0$ because a pendulum's period is a measured time, and time cannot be negative in this context. Mathematically $L = T^2$ is defined for negative $T$ (and would mirror to the left), but it has no physical meaning here.
2.1 — Explain your thinking (sample response)
The calculator's answer is actually correct because the order of operations gives powers higher priority than the unary minus — so typing $-3^2$ is read as "negative of $3$ squared", which is $-(3 \times 3) = -9$. To get $+9$, the friend needed to type $(-3)^2$, which puts the negative INSIDE the square, so it becomes $(-3) \times (-3) = +9$. The rule from Lesson 2 is: when substituting a negative number into $y = x^2$, always wrap the negative in brackets first. So for $x = -3$ we write $y = (-3)^2 = 9$ — the basic parabola sits entirely on or above the $x$-axis.
Marking: 1 mark for explaining order of operations on $-3^2$; 1 mark for the corrected input $(-3)^2$; 1 mark for stating the brackets rule clearly; 1 mark for full-sentence clarity.