Mathematics • Year 9 • Unit 2 • Lesson 4
Upside-Down Parabolas in the Real World
Use $y = -ax^2$ to model bridge arches, thrown basketballs, water fountains, and tossed paper planes — situations where the vertex is the highest point. Then explain in your own words why negative $a$ flips the parabola without moving the vertex.
1. Word problems
Each problem uses $y = -ax^2$ to describe an arc that opens downward. Show your working, with brackets around any negative substitution. 3 marks each
1.1 — Footbridge arch. A small footbridge over a creek has an arch modelled (with the top of the arch at the origin) by $y = -\tfrac{1}{4}x^2$, where $x$ is horizontal distance in metres from the centre and $y$ is the drop in metres below the centre.
(a) Find the drop $y$ at $x = 2$ m and at $x = -2$ m.
(b) State the height of the centre above the bridge's lowest landing points.
(c) Is the arch wider or narrower than the reference shape $y = -x^2$? Why?
1.2 — Basketball shot. A basketball's path (relative to its highest point as origin) is modelled by $y = -2x^2$, with $x$ in metres horizontally from the peak and $y$ in metres above the peak.
(a) Build a table for $x = -2, -1, 0, 1, 2$ metres.
(b) Sketch the parabola, marking the vertex.
(c) State the vertex's role (max or min) and explain in one sentence why that makes sense for a thrown ball.
1.3 — Water fountain jet. A garden fountain's water arc is modelled by $y = -\tfrac{1}{2}x^2$ above an origin at the fountain's peak (with $x$ in metres horizontally and $y$ in metres below the peak).
(a) Find $y$ at $x = 3$ m. How far below the peak does the water land at that horizontal distance?
(b) Is the fountain arc wider or narrower than the basketball arc in 1.2?
(c) Find the value(s) of $x$ where $y = -8$ m.
1.4 — Paper plane toss. Sam tosses a paper plane. Its height (relative to peak) follows $y = ax^2$ for some negative $a$. The plane is $1$ m below its peak when it is $2$ m horizontally away.
(a) Find $a$.
(b) Write the equation of the plane's height curve.
(c) State the direction it opens and explain why a negative $a$ is the natural choice for any thrown object.
1.5 — Comparing two arches. Two bridge arches use the equations $y = -x^2$ (low pedestrian bridge) and $y = -\tfrac{1}{5}x^2$ (long road bridge), both with the top of the arch at the origin.
(a) Find the drop $y$ at $x = 5$ metres for each arch.
(b) Which arch is wider? Justify in terms of $|a|$.
(c) Which is the more realistic shape for a long road bridge, and why?
2. Explain your thinking
Use full sentences. 4 marks
2.1 A classmate looks at $y = -3x^2$ and says: "the negative pulls the curve down by $3$, so the vertex shifts to $(0, -3)$." In your own words, explain (i) what the classmate has confused, (ii) what the negative sign actually does to the curve, and (iii) what the magnitude $3$ is doing on top of that. Use the language "reflection" and "magnitude" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Footbridge arch
(a) $y = -\tfrac{1}{4}(2)^2 = -\tfrac{1}{4} \times 4 = -1$ m. By symmetry $y = -1$ m at $x = -2$ m too.
(b) The centre sits at $y = 0$ (the top of the arch). The landing points $2$ m to each side sit at $y = -1$ m, so the centre is $\mathbf{1}$ m above the landings.
(c) $|a| = \tfrac{1}{4} < 1$, so the arch is wider than the reference $y = -x^2$ — the arms drop more slowly with horizontal distance, suitable for a longer bridge span.
1.2 — Basketball shot
(a) $x^2$: $4, 1, 0, 1, 4$. $y = -2 x^2$: $\mathbf{-8, -2, 0, -2, -8}$.
(b) Sketch: a downward U with vertex at $(0, 0)$ and symmetric points $(\pm 1, -2)$, $(\pm 2, -8)$.
(c) Vertex $(0, 0)$ is a maximum. A thrown ball reaches its highest point at the top of its arc and then falls back down on both sides, exactly matching a downward parabola with vertex at the peak.
1.3 — Water fountain jet
(a) $y = -\tfrac{1}{2}(3)^2 = -\tfrac{1}{2} \times 9 = -4.5$ m. So $\mathbf{4.5}$ m below the peak.
(b) $|{-\tfrac{1}{2}}| = 0.5$ vs basketball $|{-2}| = 2$. Smaller $|a|$ = wider. So the fountain arc is wider than the basketball arc.
(c) Set $y = -8$: $-8 = -\tfrac{1}{2}x^2 \Rightarrow x^2 = 16 \Rightarrow x = \mathbf{\pm 4}$ m. The water is $8$ m below the peak when it is $4$ m to either side.
1.4 — Paper plane toss
(a) Sub $(2, -1)$: $-1 = a(2)^2 = 4a$, so $a = \mathbf{-\tfrac{1}{4}}$.
(b) Equation: $y = -\tfrac{1}{4}x^2$.
(c) The parabola opens downward. Negative $a$ is natural for any thrown object because the peak (vertex) is the maximum height — gravity always pulls everything back below it, exactly the shape a downward parabola gives.
1.5 — Comparing two arches
(a) Pedestrian bridge $y = -x^2$ at $x = 5$: $y = -25$ m drop. Road bridge $y = -\tfrac{1}{5}x^2$ at $x = 5$: $y = -\tfrac{1}{5} \times 25 = -5$ m drop.
(b) The road bridge ($|a| = 0.2$) is the wider arch because smaller $|a|$ produces a flatter, more spread-out curve.
(c) The road bridge shape ($y = -\tfrac{1}{5}x^2$) is more realistic for a long road bridge — it stays close to horizontal across a long span (only $5$ m of drop after $5$ m horizontally), whereas $y = -x^2$ drops $25$ m in the same distance, which would be an unusable shape for traffic.
2.1 — Explain your thinking (sample response)
My classmate has confused a reflection with a translation. The negative sign in $y = -3x^2$ does not move (translate) the curve downward by any amount — it reflects the basic parabola $y = x^2$ through the $x$-axis, so the upward U becomes an upside-down U. The vertex stays exactly where it was, at $(0, 0)$, but now sits at the top instead of the bottom. The magnitude $|a| = 3$ does a separate job: it dilates the curve, making it $3$ times narrower than the basic parabola (every $y$-value is multiplied by $3$ in size). So $y = -3x^2$ is the upside-down version of $y = 3x^2$ — same vertex, same axis, just flipped.
Marking: 1 mark for "reflection, not translation"; 1 mark for "vertex stays at $(0, 0)$"; 1 mark for the magnitude job (width); 1 mark for clear, full-sentence writing.