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.