Mathematics • Year 9 • Unit 2 • Lesson 10
Sketch from a Story, Identify from a Sketch
Real situations supply you with a vertex and one extra point — your job is to find the equation, then sketch and read intercepts. Bridges, javelin throws, dive arcs, profit curves, and antenna design.
1. Word problems
For each scenario: write the equation, then answer. Show working. 3 marks each
1.1 — Suspension bridge. The cable of a small suspension bridge has its lowest point at $(20, 4)$ (m, m above the road). It also passes through the support tower at $(0, 24)$.
(a) Use $y = a(x - 20)^2 + 4$ and substitute $(0, 24)$ to find $a$.
(b) Write the full equation of the cable.
(c) Use the equation to find the cable's height above the road at $x = 30$.
1.2 — Javelin throw. A javelin's flight has its peak at $(30, 25)$ (m, m above ground) and lands at ground level at $x = 60$ m.
(a) Use the peak as the vertex: $y = a(x - 30)^2 + 25$. Sub $(60, 0)$ to find $a$.
(b) Write the equation.
(c) From the equation, find the javelin's height at $x = 10$ m. Is it higher at $x = 10$ or at $x = 45$? Justify using symmetry.
1.3 — Dive arc. A diver starts at $(0, 3)$ (the edge of a $3$ m platform), peaks at $(1, 4)$, and enters the water at some point past $x = 2$.
(a) Use the peak as the vertex: $y = a(x - 1)^2 + 4$. Sub $(0, 3)$ to find $a$.
(b) Write the equation.
(c) Use the equation to find where the diver enters the water (set $y = 0$).
1.4 — Profit curve. A small clothes shop's monthly profit (in thousands of dollars) follows the equation $P = a(n - 80)^2 + 12$, where $n$ is the number of items sold. At $n = 60$, the profit is $P = 8$ thousand.
(a) Sub $(60, 8)$ to find $a$.
(b) Write the full profit equation.
(c) The break-even point is where $P = 0$. Find the two values of $n$ where the shop breaks even.
1.5 — Antenna parabolic profile. An antenna dish has profile $y = a(x - h)^2 + k$. From the sketch, the lowest point of the dish (vertex) sits at $(0, -2)$ and the dish rim passes through $(4, 6)$.
(a) Use the vertex to fill in $h$ and $k$.
(b) Sub $(4, 6)$ to find $a$.
(c) Write the equation. Find where the dish rim meets the level $y = 0$ (i.e. find the $x$-intercepts).
2. Explain your thinking
Use full sentences, no dot points. 4 marks
2.1 A classmate is shown a parabola with vertex $(2, 3)$ and writes the equation $y = (x - 2)^2 + 3$. They don't check any other points. Write a full-paragraph response that (i) explains why ANY equation of the form $y = a(x - 2)^2 + 3$ has the right vertex (regardless of the value of $a$), (ii) explains why we still need to find $a$ before we have THE equation, (iii) describes the standard method (sub a second point and solve for $a$), and (iv) gives a concrete example of TWO different parabolas that both have vertex $(2, 3)$ but look completely different.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Suspension bridge
(a) Sub $(0, 24)$: $24 = a(0 - 20)^2 + 4 = 400a + 4 \Rightarrow 400a = 20 \Rightarrow a = \tfrac{1}{20} = 0.05$. (b) Equation: $y = 0.05(x - 20)^2 + 4$. (c) At $x = 30$: $y = 0.05(10)^2 + 4 = 0.05(100) + 4 = 5 + 4 = \mathbf{9}$ m above the road.
1.2 — Javelin throw
(a) Sub $(60, 0)$: $0 = a(60 - 30)^2 + 25 = 900a + 25 \Rightarrow 900a = -25 \Rightarrow a = -\tfrac{1}{36}$. (b) Equation: $y = -\tfrac{1}{36}(x - 30)^2 + 25$. (c) At $x = 10$: $y = -\tfrac{1}{36}(-20)^2 + 25 = -\tfrac{400}{36} + 25 \approx -11.11 + 25 \approx 13.89$ m. At $x = 45$: $y = -\tfrac{1}{36}(15)^2 + 25 = -\tfrac{225}{36} + 25 \approx -6.25 + 25 \approx 18.75$ m. Higher at $x = 45$ because it sits $15$ m from the peak, closer than $x = 10$ which is $20$ m from the peak. Distance from the axis controls height — symmetry means closer to peak = higher.
1.3 — Dive arc
(a) $y = a(x - 1)^2 + 4$. Sub $(0, 3)$: $3 = a(-1)^2 + 4 = a + 4 \Rightarrow a = -1$. (b) Equation: $y = -(x - 1)^2 + 4$. (c) Set $y = 0$: $0 = -(x - 1)^2 + 4 \Rightarrow (x - 1)^2 = 4 \Rightarrow x - 1 = \pm 2 \Rightarrow x = 3$ or $x = -1$. Take the positive root past the platform: $\mathbf{x = 3}$ m. (The diver enters the water $3$ m from the platform edge.)
1.4 — Profit curve
(a) $8 = a(60 - 80)^2 + 12 = 400a + 12 \Rightarrow 400a = -4 \Rightarrow a = -0.01 = -\tfrac{1}{100}$. (b) Equation: $P = -0.01(n - 80)^2 + 12$ (thousand dollars). (c) Set $P = 0$: $0 = -0.01(n - 80)^2 + 12 \Rightarrow (n - 80)^2 = 1200 \Rightarrow n - 80 = \pm \sqrt{1200} \approx \pm 34.64 \Rightarrow n \approx 45.4$ or $n \approx 114.6$. Break-even at approximately $\mathbf{45}$ or $\mathbf{115}$ items per month.
1.5 — Antenna parabolic profile
(a) Vertex $(0, -2)$ $\Rightarrow$ $h = 0$, $k = -2$. So $y = ax^2 - 2$. (b) Sub $(4, 6)$: $6 = 16a - 2 \Rightarrow 16a = 8 \Rightarrow a = \tfrac{1}{2}$. (c) Equation: $y = \tfrac{1}{2}x^2 - 2$. $x$-intercepts: $0 = \tfrac{1}{2}x^2 - 2 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Rim meets $y = 0$ at $(2, 0)$ and $(-2, 0)$.
2.1 — Explain your thinking (sample response)
Any equation of the form $y = a(x - 2)^2 + 3$ has vertex $(2, 3)$ because $(h, k) = (2, 3)$ regardless of the value of $a$. The vertex sits where the bracket is zero (here $x = 2$), and at that $x$ the $y$-value is just $k = 3$. So my classmate hasn't done anything wrong with the vertex — but they have stopped too early. Different values of $a$ give parabolas that all SHARE that vertex but have completely different shapes — some narrow, some wide, some opening up, some opening down. So we need a SECOND piece of information to pin down $a$. The standard method is: substitute one more point (any point on the parabola other than the vertex) into $y = a(x - 2)^2 + 3$ and solve the resulting equation for $a$. For a concrete example, two parabolas that share vertex $(2, 3)$ but look very different are $y = (x - 2)^2 + 3$ (a standard upward U with $y$-intercept $(0, 7)$) and $y = -3(x - 2)^2 + 3$ (a narrow downward inverted U with $y$-intercept $(0, -9)$). Same vertex; very different curves.
Marking: 1 mark for "any $a$ gives the same vertex"; 1 mark for "still need $a$"; 1 mark for sub-and-solve method; 1 mark for concrete contrasting example.