Mathematics • Year 9 • Unit 2 • Lesson 3
Stretching and Squashing Parabolas
Use $y = ax^2$ to model satellite dishes, soccer kicks, fountain jets, designer skate ramps and headlight reflectors. Then explain in your own words why a fraction $a$ makes the curve wider, not narrower.
1. Word problems
Each problem uses $y = ax^2$ to describe something physical. Show your working, including the "square first, then multiply" order. 3 marks each
1.1 — Satellite dish profile. A small satellite dish has cross-section $y = \tfrac{1}{4}x^2$ (where $x$ is horizontal distance in metres from the centre, and $y$ is the dish depth in metres). It needs to reach a depth of $y = 1$ m at its edge.
(a) Find the value(s) of $x$ where $y = 1$.
(b) State the total width of the dish (edge to edge).
(c) Is this dish wider or narrower than the reference curve $y = x^2$? One-sentence reason.
1.2 — Soccer ball table. A child kicks a soccer ball straight up. Its height above the start point (in metres) at time $t$ (in seconds before the ball reaches its peak) is modelled by $h = 5t^2$ (taking $t = 0$ as the peak, then $t$ counts down to the moment of release).
(a) Build a table for $t = 0, 1, 2$.
(b) Compared to the reference $y = x^2$, is the soccer-ball curve wider or narrower? Why?
(c) Find $t$ when $h = 20$ metres.
1.3 — Fountain water arch. A garden fountain shoots water from a central nozzle. The water's path on each side is modelled by $y = -\tfrac{1}{8}x^2 + 2$ for the upward arc, but a designer is approximating it just with the dilation part: $y = \tfrac{1}{8}x^2$ (for the falling cone shape).
(a) For the dilation $y = \tfrac{1}{8}x^2$, find $y$ when $x = 4$.
(b) Find $y$ when $x = 8$ — how many times bigger than your $x = 4$ answer is it?
(c) Is $y = \tfrac{1}{8}x^2$ wider or narrower than $y = x^2$?
1.4 — Designer skate ramp. A skate ramp's cross-section is modelled by $y = ax^2$. The designer wants the ramp to be $1$ metre high when $x = 2$ metres from the centre line.
(a) Find $a$ so that the ramp passes through $(2, 1)$.
(b) Write the ramp's equation.
(c) Compared to a standard $y = x^2$ ramp, is this one wider or narrower? Why does that make it safer for beginners?
1.5 — Headlight reflector. A car's parabolic headlight reflector has cross-section $y = 3x^2$ (with $x$ in cm). Compare it with a wider truck headlight $y = \tfrac{1}{2}x^2$ at the same horizontal distance $x = 2$ cm.
(a) Compute $y$ for both reflectors at $x = 2$ cm.
(b) Which reflector is deeper at that horizontal distance, and by how many cm?
(c) Which reflector's curve is narrower — the car or the truck — and what does that mean for how tightly the headlight beam focuses?
2. Explain your thinking
Use full sentences. 4 marks
2.1 A classmate says "$y = \tfrac{1}{2}x^2$ must be narrower than $y = x^2$ because $\tfrac{1}{2}$ is a small number, and small things are skinny." In your own words, explain (i) what the classmate has confused, (ii) what $a = \tfrac{1}{2}$ actually does to every $y$-value compared with $y = x^2$, and (iii) why halving the $y$-values produces a WIDER curve, not a narrower one.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Satellite dish
(a) Set $y = 1$: $1 = \tfrac{1}{4}x^2 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. So the edges are at $x = -2$ m and $x = 2$ m.
(b) Total width: from $-2$ to $+2$ is $\mathbf{4}$ m.
(c) $a = \tfrac{1}{4} < 1$, so this dish is wider than the reference $y = x^2$ — the curve spreads out more for the same depth.
1.2 — Soccer ball table
(a) $t = 0$: $h = 0$. $t = 1$: $h = 5$. $t = 2$: $h = 20$.
(b) $a = 5 > 1$, so narrower than $y = x^2$. The ball gains height fast as $t$ increases (because gravity is fairly strong in the model).
(c) $20 = 5t^2 \Rightarrow t^2 = 4 \Rightarrow t = 2$ s (positive root, since time can't be negative).
1.3 — Fountain water arch
(a) $y = \tfrac{1}{8}(4)^2 = \tfrac{16}{8} = \mathbf{2}$ m.
(b) $y = \tfrac{1}{8}(8)^2 = \tfrac{64}{8} = \mathbf{8}$ m. That's $\dfrac{8}{2} = 4$ times bigger — matches our prediction that doubling $x$ quadruples $y$.
(c) $a = \tfrac{1}{8} < 1$, so the curve is wider than $y = x^2$.
1.4 — Designer skate ramp
(a) Sub $(2, 1)$: $1 = a(2)^2 = 4a$, so $a = \mathbf{\tfrac{1}{4}}$.
(b) Equation: $y = \tfrac{1}{4}x^2$.
(c) $a = \tfrac{1}{4} < 1$, so the ramp curve is wider than $y = x^2$ — the rise is more gradual for any given horizontal distance, meaning the ramp is less steep, which is safer for beginners learning balance.
1.5 — Headlight reflector
(a) Car: $y = 3(2)^2 = 12$ cm. Truck: $y = \tfrac{1}{2}(2)^2 = 2$ cm.
(b) The car reflector is deeper, by $12 - 2 = \mathbf{10}$ cm at $x = 2$ cm.
(c) The car reflector is narrower ($a = 3 > 1$). A narrower parabola focuses incoming light to a tighter point (closer to the focus), giving a more concentrated beam — useful for a car headlight aimed straight down the road.
2.1 — Explain your thinking (sample response)
My classmate has confused the size of the coefficient $a$ with the appearance of the curve — "small number" does not mean "skinny curve". In $y = ax^2$ the value of $a$ is a multiplier applied to every $y$-value. When $a = \tfrac{1}{2}$, every $y$-value of $y = x^2$ is halved — for example at $x = 2$, $y = x^2$ gives $y = 4$, but $y = \tfrac{1}{2}x^2$ gives only $y = 2$. Halving the heights flattens the curve closer to the $x$-axis, which makes its arms spread out further from the $y$-axis before they reach any given height — producing a wider parabola, not a narrower one. The rule from Lesson 3 is: bigger $a$ means narrower; smaller positive $a$ (between $0$ and $1$) means wider.
Marking: 1 mark for naming the confusion; 1 mark for "$a$ multiplies every $y$-value"; 1 mark for the wider-not-narrower conclusion; 1 mark for clarity and a worked example.