Mathematics • Year 9 • Unit 2 • Lesson 5
Sliding the Parabola Up and Down
Use $y = x^2 + c$ to model thermometer offsets, sea-floor profiles, garden pond depths, lifting platforms, and elevation maps. Then explain why adding a constant shifts the curve vertically rather than sideways.
1. Word problems
Each problem uses $y = x^2 + c$ as a model. Show your working — including testing whether $x$-intercepts make sense in the context. 3 marks each
1.1 — Pond depth profile. A garden pond's cross-section is modelled by $y = x^2 - 4$ (where $x$ is horizontal distance in metres from the centre and $y$ is the depth in metres relative to ground level — negative $y$ means below ground).
(a) Find the deepest point and state where (which $x$) it occurs.
(b) Find the $x$-intercepts of the model and interpret them as the pond's edges.
(c) Compute the pond's total width (edge to edge).
1.2 — Thermometer offset. A weather logger records temperatures relative to a baseline. The plotted curve has form $T = x^2 + 5$, where $x$ is hours from midday (negative = morning, positive = afternoon) and $T$ is temperature in $^\circ$C.
(a) Find $T$ at $x = -2$ and at $x = 2$. What does that tell you about the model's symmetry?
(b) State the minimum temperature predicted by the model and when it occurs.
(c) Are there any times when $T = 0\,^\circ$C? Show why or why not.
1.3 — Lifted platform. A small skateboard launch platform has cross-section $y = x^2 + 1$ (with $x$ in metres from the centre and $y$ in metres above ground).
(a) Find the height at $x = 0$ and at $x = 2$.
(b) State the vertex of the platform shape.
(c) Compare with $y = x^2$ — what does "$+1$" do to every $y$-value, and what does it do to the vertex?
1.4 — Sea-floor dip. A coastal survey models a small sea-floor dip by $y = x^2 - 9$ (with $x$ in metres from the dip's centre and $y$ in metres below sea level — negative $y$ means below).
(a) State the deepest point.
(b) Find the $x$-intercepts and interpret them as where the dip rejoins the flat sea floor at $y = 0$.
(c) Build a small table for $x = -3, -2, -1, 0, 1, 2, 3$ to sketch the dip.
1.5 — Hilltop lookout. A hill's elevation profile is approximated by $y = -x^2 + 25$ (downward parabola: $a = -1$, $c = 25$). This is a small extension of Lesson 5 — combining reflection (L4) with vertical translation (L5).
(a) State the highest point (vertex) and the height at $x = 3$ m.
(b) Find the $x$-intercepts — where the hill meets ground level $y = 0$.
(c) Compute the total width of the hill (edge to edge at ground level).
2. Explain your thinking
Use full sentences. 4 marks
2.1 A classmate looks at $y = x^2 + 3$ and says "the $+3$ shifts the parabola $3$ to the right." In your own words, explain (i) what the $+3$ actually does to the curve, (ii) what the vertex becomes, and (iii) which kind of equation WOULD shift it sideways. Use the words "vertical" and "vertex" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Pond depth profile
(a) Deepest point is the vertex $(0, -4)$ — $4$ m below ground, at $x = 0$ (the centre).
(b) Set $y = 0$: $x^2 - 4 = 0 \Rightarrow x = \pm 2$. The edges of the pond (where it meets ground level) are at $x = -2$ m and $x = 2$ m.
(c) Total width: $2 - (-2) = \mathbf{4}$ m.
1.2 — Thermometer offset
(a) $T(-2) = (-2)^2 + 5 = 4 + 5 = 9\,^\circ$C. $T(2) = (2)^2 + 5 = 4 + 5 = 9\,^\circ$C. Same value — the model is symmetric about $x = 0$ (midday).
(b) Minimum $T$ is the vertex value: $T = 5\,^\circ$C, at $x = 0$ (midday).
(c) Set $T = 0$: $x^2 + 5 = 0 \Rightarrow x^2 = -5$. No real solutions — the model never reaches $0\,^\circ$C, because the vertex sits at $5\,^\circ$C above the $x$-axis and the curve only goes up from there.
1.3 — Lifted platform
(a) At $x = 0$: $y = 0 + 1 = 1$ m. At $x = 2$: $y = 4 + 1 = 5$ m.
(b) Vertex: $(0, 1)$.
(c) Adding $1$ raises every $y$-value of $y = x^2$ by $1$ — the whole curve slides up by $1$ unit. The vertex shifts from $(0, 0)$ to $(0, 1)$. The shape (width, direction) is unchanged.
1.4 — Sea-floor dip
(a) Deepest point is the vertex $(0, -9)$ — $9$ m below sea level at the dip's centre.
(b) Set $y = 0$: $x^2 - 9 = 0 \Rightarrow x = \pm 3$ m. So the dip rejoins flat sea floor at $x = -3$ m and $x = 3$ m.
(c) Table: $y = x^2 - 9$ at $x = -3, -2, -1, 0, 1, 2, 3$ gives $y = 0, -5, -8, -9, -8, -5, 0$.
1.5 — Hilltop lookout
(a) Vertex (highest point): $(0, 25)$, so the hill is $25$ m tall at its centre. At $x = 3$: $y = -(3)^2 + 25 = -9 + 25 = 16$ m.
(b) Set $y = 0$: $0 = -x^2 + 25 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$ m.
(c) Total width: from $x = -5$ to $x = 5$, width $= \mathbf{10}$ m.
2.1 — Explain your thinking (sample response)
My classmate has confused a vertical shift with a horizontal one. In $y = x^2 + 3$ the $+3$ is added to the OUTPUT $y$, not to the input $x$, so it raises every $y$-value by $3$ — the whole curve slides $3$ units straight UP, not sideways. The vertex moves from $(0, 0)$ on $y = x^2$ to $(0, 3)$ on $y = x^2 + 3$, but the axis of symmetry stays at $x = 0$. To actually shift a parabola sideways (e.g. $3$ to the right), the equation would have to subtract from $x$ inside the brackets: $y = (x - 3)^2$ — you'll meet that form in a later lesson, but it's not what's happening here.
Marking: 1 mark for "vertical, not horizontal"; 1 mark for "vertex moves to $(0, 3)$"; 1 mark for naming $(x - h)^2$ as the sideways form; 1 mark for clarity and use of the required words.