Mathematics • Year 9 • Unit 2 • Lesson 9
Reading Intercepts in Real Situations
Use intercepts to answer real questions: when does a thrown ball hit the ground, where does an arch meet the floor, when does profit hit zero, what's the starting price, and where does a tunnel meet the road.
1. Word problems
For each scenario: find the intercept you need, answer the question, and show working. 3 marks each
1.1 — Cricket ball. A cricket ball is hit from ground level and follows the path $y = -x^2 + 8x$, where $x$ is horizontal distance (m) and $y$ is height (m).
(a) Find the $y$-intercept and explain what it means physically.
(b) Find the $x$-intercepts (factor out $x$). What do they tell you?
(c) How far away does the ball land from where it was hit?
1.2 — Stone arch. A stone-arch entrance has cross-section $y = -(x - 3)^2 + 9$, where $x$ is the horizontal distance (m) across the entrance and $y$ is the height (m) above the floor.
(a) Find the $y$-intercept and say what it means at the start of the arch ($x = 0$).
(b) Find the $x$-intercepts — where does the arch meet the floor?
(c) How wide is the entrance at floor level?
1.3 — Break-even point. A coffee cart's daily profit is $P = -(n - 50)^2 + 400$, where $n$ is the number of coffees sold and $P$ is profit in dollars.
(a) Find the $P$-intercept (i.e. sub $n = 0$) and explain what it means.
(b) Find the $n$-values where $P = 0$ (the "break-even points") by setting $P = 0$.
(c) Between which two numbers of coffees does the cart make a profit?
1.4 — Tunnel meets road. A road tunnel has cross-section $y = -\tfrac{1}{2}(x - 4)^2 + 8$, where $x$ is the width (m) across the tunnel from the left wall base and $y$ is the height (m) of the tunnel ceiling above the road.
(a) Find the $x$-intercepts to see where the tunnel walls meet the road.
(b) How wide is the tunnel at road level?
(c) How tall is the tunnel at the centre? (Hint: use the vertex.)
1.5 — Will it hit the ground? Three parabolic paths are proposed for a thrown frisbee. Predict (WITHOUT solving algebraically) which ones return to ground level ($y = 0$) and which never do. Justify each.
(a) $y = -(x - 3)^2 + 5$
(b) $y = 2(x - 2)^2 + 1$
(c) $y = -\tfrac{1}{4}(x - 6)^2 + 0$ (i.e. $y = -\tfrac{1}{4}(x - 6)^2$).
2. Explain your thinking
Use full sentences, no dot points. 4 marks
2.1 A classmate is asked "Find the $x$-intercepts of $y = x^2 + 9$." They write: "Set $y = 0$: $x^2 + 9 = 0 \Rightarrow x^2 = -9 \Rightarrow x = \pm 3$." Write a full-paragraph reply that (i) identifies the specific arithmetic mistake (where $-9 \to \pm 3$ goes wrong), (ii) explains why $x^2$ can NEVER equal a negative number for real $x$, (iii) describes how the lesson's vertex+direction shortcut would have caught this immediately, and (iv) writes the correct conclusion for the question.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Cricket ball
(a) $y$-intercept: sub $x = 0$, $y = 0$. So $(0, 0)$ — the ball starts at ground level (the batting point). (b) Factor: $-x^2 + 8x = -x(x - 8) = 0$ when $x = 0$ or $x = 8$. So $x$-intercepts $(0, 0)$ and $(8, 0)$ — these are the launch point and the landing point. (c) The ball lands $\mathbf{8}$ m from where it was hit.
1.2 — Stone arch
(a) $y$-intercept: $y = -(0 - 3)^2 + 9 = -9 + 9 = 0$, so $(0, 0)$. This means the left side of the arch meets the floor exactly at $x = 0$. (b) $x$-intercepts: $0 = -(x - 3)^2 + 9 \Rightarrow (x - 3)^2 = 9 \Rightarrow x - 3 = \pm 3 \Rightarrow x = 6$ or $x = 0$. Points $(0, 0)$ and $(6, 0)$. (c) Width at floor: $6 - 0 = \mathbf{6}$ m.
1.3 — Break-even point
(a) Sub $n = 0$: $P = -(-50)^2 + 400 = -2500 + 400 = -\$2100$. Loss of $\$2100$ if no coffees are sold (fixed costs). (b) Set $P = 0$: $-(n - 50)^2 + 400 = 0 \Rightarrow (n - 50)^2 = 400 \Rightarrow n - 50 = \pm 20 \Rightarrow n = 70$ or $n = 30$. Break-even: $\mathbf{30}$ or $\mathbf{70}$ coffees. (c) Profit between $n = 30$ and $n = 70$ — i.e. selling somewhere from 31 to 69 coffees yields a positive profit.
1.4 — Tunnel meets road
(a) $0 = -\tfrac{1}{2}(x - 4)^2 + 8 \Rightarrow (x - 4)^2 = 16 \Rightarrow x - 4 = \pm 4 \Rightarrow x = 0$ or $x = 8$. So the walls meet the road at $(0, 0)$ and $(8, 0)$. (b) Width at road level: $8 - 0 = \mathbf{8}$ m. (c) Centre height = $k$-value = $\mathbf{8}$ m (vertex $(4, 8)$).
1.5 — Will it hit the ground?
(a) Vertex $(3, 5)$ ABOVE $x$-axis; opens DOWN ($a = -1$). The curve must come back down through the $x$-axis: YES, hits the ground (2 intercepts).
(b) Vertex $(2, 1)$ ABOVE $x$-axis; opens UP ($a = 2$). Already above, going further up: NO, never reaches the ground (0 intercepts).
(c) Vertex $(6, 0)$ exactly ON the $x$-axis. Touches at one point only: YES, but only just (1 repeated intercept at $x = 6$).
2.1 — Explain your thinking (sample response)
The arithmetic mistake happens at the step "$x^2 = -9 \Rightarrow x = \pm 3$". The student has correctly isolated $x^2$ as $-9$, but then they have ignored the negative sign and pretended $\sqrt{9} = \pm 3$ works. The right square root of $-9$ is NOT $\pm 3$ — there is no real number whose square is $-9$, because squaring any real number (positive, negative, or zero) always produces a result $\ge 0$. To see this, $(+3)^2 = 9$ and $(-3)^2 = 9$ — both positive. So $x^2 = -9$ has NO real solutions, and the curve $y = x^2 + 9$ never crosses the $x$-axis. The lesson's vertex+direction shortcut would have caught this immediately: vertex is $(0, 9)$, which sits ABOVE the $x$-axis, and the curve opens UP ($a = 1 > 0$), so the curve stays above the $x$-axis everywhere — zero $x$-intercepts. The correct conclusion is: $y = x^2 + 9$ has NO real $x$-intercepts.
Marking: 1 mark for spotting the $x^2 = -9 \to \pm 3$ step; 1 mark for "$x^2$ can't be negative"; 1 mark for the vertex+direction check; 1 mark for the correct conclusion.