Mathematics • Year 9 • Unit 2 • Lesson 16
Choosing Scale and Shape — Real Graphs
Five real scenarios that need a careful graph: roller-coaster height, bird-watching populations, tile-pattern area, share price growth, road-trip fuel. Pick the right $x$-values, the right scale, and join with the right curve.
1. Word problems
For each: identify the family, choose $x$-values, build the table, pick a sensible scale, then describe the shape you'd plot. 3 marks each
1.1 — Roller-coaster dip. The height of a roller-coaster car above the ground is modelled (over one section) by $h = t^2 - 6t + 5$ (m, s), for $t = 0$ to $t = 6$ seconds.
(a) Build a table for $t = 0, 1, 2, 3, 4, 5, 6$.
(b) Name the family and state the lowest point (the dip).
(c) Suggest a sensible $y$-scale (units per square) given the $y$-range.
1.2 — Bird-watching ratio. A bird-watching group spots a total of $24$ birds and divides them equally among $n$ observers. Birds per observer is $B = \dfrac{24}{n}$.
(a) Build a table of $B$ for $n = 1, 2, 3, 4, 6, 8, 12, 24$.
(b) Name the family and explain why we DON'T plot $n = 0$.
(c) Describe the shape of the graph in one sentence.
1.3 — Tile-pattern area. A square tile pattern has $L$ tiles per side. The total number of tiles is $T = L^2$.
(a) Build a table for $L = 1, 2, 3, 4, 5, 6$.
(b) Name the family.
(c) The page has only $36$ vertical squares on the grid. Suggest a sensible $y$-scale so the graph fits.
1.4 — Share price growth. A share starts at $\$100$ and grows by a factor of $1.5$ each year, so $P = 100 \times 1.5^t$ for whole-year $t = 0, 1, 2, 3, 4$.
(a) Build the table of share prices for $t = 0$ to $t = 4$ (round to the nearest dollar).
(b) Name the family.
(c) State the equation of the horizontal asymptote (the long-run lower limit if you extended $t$ very negative).
1.5 — Road-trip fuel. A $480$ km road trip's total time (in hours) depends on the average speed $v$ (km/h) used: $T = \dfrac{480}{v}$.
(a) Build the table of $T$ for $v = 40, 60, 80, 96, 120, 160$ km/h.
(b) Name the family.
(c) Why is the realistic domain $v > 0$ only? What does that mean for the shape of the graph?
2. Explain your thinking
Use full sentences. 4 marks
2.1 A classmate plots $y = x^2$ using only $x = 0, 1, 2, 3$ and connects the four points with straight line segments. In your own words, explain (i) why their choice of $x$-values misses half the parabola, (ii) why "connect with straight segments" is wrong for any non-linear graph, and (iii) what they should do differently to get the correct picture.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Roller-coaster dip
(a) $t = 0, 1, 2, 3, 4, 5, 6$ gives $h = 5, 0, -3, -4, -3, 0, 5$.
(b) Family: parabola. Lowest point (the dip): vertex at $(3, -4)$ — $4$ m below ground level at $t = 3$ s.
(c) $y$-range $-4$ to $5$: $1$ unit per square works (9 vertical squares total). The interpretation note: if "ground level" is $h = 0$, then negative $h$ would mean a tunnel section.
1.2 — Bird-watching ratio
(a) $n = 1, 2, 3, 4, 6, 8, 12, 24$ gives $B = 24, 12, 8, 6, 4, 3, 2, 1$.
(b) Family: hyperbola. We don't plot $n = 0$ because $\tfrac{24}{0}$ is undefined — you can't divide birds among zero people.
(c) A single smooth curve in Quadrant 1, falling quickly at first ($B$ drops from $24$ to $4$ as $n$ goes $1$ to $6$), then more slowly ($B$ drops from $4$ to $1$ as $n$ goes $6$ to $24$).
1.3 — Tile-pattern area
(a) $L = 1, 2, 3, 4, 5, 6$ gives $T = 1, 4, 9, 16, 25, 36$.
(b) Family: parabola (right-hand arm only since $L \ge 1$).
(c) Max $T = 36$, with $36$ vertical squares available: $1$ unit per square just fits. (If the grid were smaller, use $2$ units per square to give yourself headroom.)
1.4 — Share price growth
(a) $t = 0, 1, 2, 3, 4$ gives $P \approx 100, 150, 225, 338, 506$ (rounded to nearest dollar).
(b) Family: exponential growth (base $1.5 > 1$).
(c) Horizontal asymptote: $P = 0$ — for very negative $t$ (long before the share existed), $P$ approaches but never touches $\$0$.
1.5 — Road-trip fuel
(a) $v = 40, 60, 80, 96, 120, 160$ gives $T = 12, 8, 6, 5, 4, 3$ hours.
(b) Family: hyperbola ($T = \tfrac{480}{v}$, $k = 480$).
(c) Negative speed isn't physical, so only $v > 0$ makes sense. That means only ONE branch (in Quadrant 1) appears in the graph — not two. The $T$-axis is a vertical asymptote (as $v \to 0$, $T \to \infty$); the $v$-axis is a horizontal asymptote (as $v \to \infty$, $T \to 0$).
2.1 — Explain your thinking (sample response)
My classmate is wrong in two ways. First, choosing only positive $x$-values ($0, 1, 2, 3$) misses the LEFT arm of the parabola: $y = x^2$ is symmetric across the $y$-axis, so values like $x = -1, -2, -3$ give the mirror $y$-values ($1, 4, 9$) and complete the U-shape. By only plotting one side, they end up with half a curve that looks more like a curved line than a parabola. Second, "straight line segments" between points is wrong because the very word non-linear means "not a straight line". A parabola has a continuous, smoothly changing slope — between any two plotted points it curves, not zig-zags. What they should do is use $x = -3, -2, -1, 0, 1, 2, 3$ for symmetry, compute all seven $y$-values, plot all seven points, then join them with a SINGLE smooth U-shape (no kinks, no straight bits).
Marking: 1 mark for "missing the negative side / symmetry"; 1 mark for "smooth, not straight"; 1 mark for the correct fix (more symmetric $x$-values + smooth curve); 1 mark for clear, full-sentence writing.