Mathematics • Year 9 • Unit 2 • Lesson 6
Parabolas in the Wild: $y = ax^2 + c$
Use $y = ax^2 + c$ for real situations: a dropped object, a satellite dish, a thrown ball, a fountain arc, and bridge geometry. Read $a$ and $c$ off each scenario, then explain in plain Year 9 English.
1. Word problems
For each scenario: identify $a$ and $c$, then answer the questions. Show working. 3 marks each
1.1 — Dropped from a bridge. A stone is dropped from a bridge $20$ m above a river. Its height above the river after $t$ seconds is modelled by $h = -5t^2 + 20$ (height in metres, $t$ in seconds).
(a) State the values of $a$ and $c$ in this $y = ax^2 + c$ form.
(b) State the vertex of the curve. What does the vertex mean in the real world?
(c) When does the stone hit the water? (Set $h = 0$ and solve.)
1.2 — Satellite dish cross-section. The cross-section of a small satellite dish has equation $y = \tfrac{1}{4}x^2 - 1$, with $x$ in metres across the dish and $y$ the height in metres at that point.
(a) Does the dish open upward or downward? How can you tell from the equation?
(b) State the lowest point of the dish (the vertex).
(c) Find the points where the dish meets the rim ($y = 0$). How wide is the dish across the top?
1.3 — Cliff diver. A diver jumps off a $10$ m cliff. Her height above the water is modelled by $h = -5t^2 + 10$ (metres, seconds).
(a) From the equation, state her starting height (the vertex value $c$).
(b) Without computing, will she hit the water? Justify using the signs of $a$ and $c$.
(c) Find when she hits the water (solve $h = 0$, positive $t$ only).
1.4 — Garden fountain. A garden fountain shoots water in a parabolic arc. Its height (m) above the spout at horizontal distance $x$ (m) is $y = -2x^2 + 3$.
(a) State the maximum height of the water and where it occurs.
(b) How far from directly above the spout does the water hit the ground ($y = 0$)? Give the answer to 2 decimal places.
(c) Across the whole arc (left and right of the spout), how wide is the fountain at ground level?
1.5 — Stretched vs unstretched arch. Bridge A has the cross-section $y = -x^2 + 9$. Bridge B has $y = -2x^2 + 9$ (both in metres, $y$ = height above the road).
(a) Which bridge has the taller centre (the higher vertex)? Justify in one sentence.
(b) For each bridge, find the $x$-intercepts — these tell you the width at road level.
(c) Which bridge is wider at road level, and by how many metres?
2. Explain your thinking
Use full sentences, no dot points. 4 marks
2.1 A classmate is sketching $y = -2x^2 + 5$ and writes that the vertex is at $(0, -2)$ because "the number in front of $x^2$ is the vertex height." Write a full-paragraph explanation that (i) names the actual vertex of the curve, (ii) clearly explains which letter ($a$ or $c$) sets the vertex height for $y = ax^2 + c$ and which letter does something else, (iii) describes what the $-2$ in this equation actually controls, and (iv) gives a one-line check the classmate could use next time to avoid this mistake.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Dropped from a bridge
(a) $a = -5$, $c = 20$. (b) Vertex $(0, 20)$ — at $t = 0$ the stone is at the top of its drop ($20$ m up); this is its maximum height. (c) Set $h = 0$: $-5t^2 + 20 = 0 \Rightarrow t^2 = 4 \Rightarrow t = \pm 2$. Take the positive root: $\mathbf{t = 2}$ seconds.
1.2 — Satellite dish cross-section
(a) Opens UPWARD because $a = \tfrac{1}{4} > 0$. (b) Vertex $(0, -1)$ — the lowest point of the dish is $1$ m below the rim level $y = 0$. (c) Set $y = 0$: $\tfrac{1}{4}x^2 - 1 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Width across the top: $2 - (-2) = \mathbf{4}$ m.
1.3 — Cliff diver
(a) Starting height $c = \mathbf{10}$ m (the vertex is at $(0, 10)$). (b) Yes, she hits the water. $a = -5 < 0$ and $c = 10 > 0$ have opposite signs, so the curve definitely crosses the $x$-axis (the water line) twice. (c) $-5t^2 + 10 = 0 \Rightarrow t^2 = 2 \Rightarrow t = \pm\sqrt{2} \approx \pm 1.41$. Take the positive root: $\mathbf{t \approx 1.41}$ seconds.
1.4 — Garden fountain
(a) Max height = vertex $y$-value $= c = \mathbf{3}$ m, at $x = 0$ (directly above the spout). (b) Set $y = 0$: $-2x^2 + 3 = 0 \Rightarrow x^2 = \tfrac{3}{2} \Rightarrow x = \pm\sqrt{1.5} \approx \pm 1.22$ m. So the water hits the ground $\approx 1.22$ m from directly above the spout. (c) Width = $1.22 - (-1.22) = \mathbf{2.45}$ m (to 2 dp).
1.5 — Stretched vs unstretched arch
(a) Same centre height: both have $c = 9$, so both have vertex $(0, 9)$ — they are equally tall in the middle. (b) Bridge A: $-x^2 + 9 = 0 \Rightarrow x = \pm 3$ (width 6 m). Bridge B: $-2x^2 + 9 = 0 \Rightarrow x^2 = 4.5 \Rightarrow x \approx \pm 2.12$ (width $\approx 4.24$ m). (c) Bridge A is wider at road level by $6 - 4.24 \approx \mathbf{1.76}$ m. (Larger $|a|$ makes a narrower parabola.)
2.1 — Explain your thinking (sample response)
The vertex of $y = -2x^2 + 5$ is actually $(0, 5)$, not $(0, -2)$. In the general form $y = ax^2 + c$, the letter $\mathbf{c}$ is what sets the vertex height — it is the constant on the end of the equation, not the coefficient in front of $x^2$. The letter $\mathbf{a}$ does something completely different: its sign controls which way the parabola opens (positive = up, negative = down) and its size controls the width (bigger $|a|$ = narrower curve). So in $y = -2x^2 + 5$, the $-2$ is $a$: it means the parabola opens DOWNWARD and is NARROWER than $y = x^2$. The $+5$ on the end is $c$, and that is the vertex height. A one-line check: substitute $x = 0$. Whatever $y$-value you get is $c$, and that is the vertex height.
Marking: 1 mark for correct vertex $(0, 5)$; 1 mark for clearly separating the jobs of $a$ and $c$; 1 mark for describing what the $-2$ actually does; 1 mark for the substitution check.