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hscscienceMaths Std · Y12
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Algebraic Relationships · L5 of 6 ~50 min MST-12-S2-01 ⚡ +90 XP available

Quadratic Models, Applying and Interpreting

Reading a graph and answering questions is only half the skill. The other half is interpreting what the features mean in the real world, and being critical enough to say when the model stops making sense.

Today's hook, A farmer has 80 m of fencing to enclose a rectangular paddock. If the area is $A = 40w - w^2$, what width $w$ gives the most area, and is a width of 50 m even physically possible?
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Recall, your gut answer first
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A farmer has 80 m of fencing to enclose a rectangular paddock. One side is the barn wall (free). The area is modelled by $A = 40w - w^2$ where $w$ is the width of the paddock in metres.

Without calculating: What width do you think gives the maximum area? And can a width of $w = 50$ m be valid, why or why not?

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The key idea you need to own
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A quadratic model only makes physical sense within its valid domainthe range of $x$ values for which the model gives meaningful, real-world results. Beyond the valid domain, the model may predict negative areas, imaginary times, or impossible quantities.

The valid domain is typically bounded by the $x$-intercepts (where the quantity is zero) and any practical physical constraints. The model is only valid between these limits.

Valid domain: usually $x_1 \leq x \leq x_2$ where $x_1, x_2$ are the $x$-intercepts (or physical limits)
Negative values may be invalid
If the model gives a negative area, height, or distance, it has left the valid domain. The physical constraint (e.g. "height can't be negative") defines the boundary.
Context drives interpretation
The vertex, $x$-intercepts, and $y$-intercept each have specific meanings. Always state the meaning in the language of the problem, not just the numbers.
Model limitations
A quadratic model assumes a simple symmetric structure. Real situations are rarely perfectly quadratic, state what the model assumes and where it might diverge from reality.
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What you'll master
Know

Key facts

  • The valid domain of a quadratic model is bounded by real-world constraints
  • $x$-intercepts, vertex and $y$-intercept each have specific contextual meanings
  • A quadratic model is only valid where it gives physically meaningful results
  • Limitations arise because the model assumes a fixed, symmetric quadratic shape
Understand

Concepts

  • How to use a given quadratic graph to answer practical questions
  • Why the valid domain is usually confined to between the $x$-intercepts
  • What the vertex means in different contexts (max height, max area, max profit)
  • Why the quadratic model has limitations in real-world applications
Can do

Skills

  • Read a quadratic graph and answer contextual questions about a practical situation
  • Interpret the meaning of each intercept and the vertex in context
  • State the valid domain with both the inequality and its contextual meaning
  • Identify and explain at least two limitations of a quadratic model in a given context
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Key terms
Valid domainThe set of $x$-values for which the quadratic model gives physically meaningful (realistic) results. Determined by real-world constraints, often between the $x$-intercepts.
Model limitationA condition under which the model breaks down or gives unrealistic predictions, e.g. predicting negative heights, speeds beyond physical limits.
Practical maximum/minimumThe vertex value interpreted in context: maximum height, maximum area, minimum cost, etc., the real-world significance of the turning point.
Model validityWhether the equation appropriately represents the real situation. A model is valid when its assumptions match the real behaviour within the domain of interest.
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Interpreting Intercepts and Vertex in Context
core concept

Every feature of a parabola has a specific real-world meaning that depends entirely on what the variables represent. The interpretation is never just the coordinates, it is the story those coordinates tell.

What each feature tells you:

  • $y$-intercept: The value of the modelled quantity when $x = 0$, the initial or starting value. E.g. if $x$ = time, this is the quantity at time zero. If $x$ = units produced, this is the fixed cost with zero production.
  • $x$-intercepts: Where the quantity is zero. E.g. when height = 0 (ball at ground), when profit = 0 (break-even), when area = 0 (no paddock). There may be two $x$-intercepts bounding the practical range.
  • Vertex: The maximum (opens down) or minimum (opens up). Always the most important point: maximum height reached, maximum area enclosed, minimum cost, etc. State both the $x$ and $y$ coordinates with units and meaning.
Farmer paddock example: $A = 40w - w^2$. $y$-intercept $(0, 0)$: zero area when width is zero, makes sense. $x$-intercepts: $w = 0$ and $w = 40$, area is zero when width is zero or 40 m (all fence used in width, none left for length). Vertex: $w = 20$ m, $A = 40(20) - 400 = 400$ m², maximum area of 400 m² at width 20 m.

The y-intercept gives the initial value (when x = 0); the x-intercepts give the zeros of the model; the vertex gives the maximum or minimum output value and the input at which it occurs.

Pause, copy the three contextual meanings: y-intercept = initial value when x = 0, x-intercepts = input values where the output is zero, vertex = maximum or minimum output value, with a real example for each.

Quick check: For a ball thrown upward with height $h = -5t^2 + 30t$ (metres, seconds), what does the second $x$-intercept represent?

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Valid Domain and Model Limitations
core concept

The y-intercept gives the starting value, x-intercepts give the zeros, and the vertex gives the optimal value in context, but a parabola extends infinitely in both directions, while real situations do not. The valid domain limits the model: time can't be negative, a ball can't be below the ground, and a length can't exceed the available material.

A mathematical model produces output for any $x$-value, but only a restricted range gives physically meaningful results. Identifying the valid domain is a critical thinking skill, NESA tests it explicitly.

How to determine the valid domain:

  1. Identify the physical quantity on the $x$-axis (time, width, units produced, etc.)
  2. Determine what values are physically possible negative time is impossible; a paddock width can't exceed the total fence; a price can't be negative.
  3. Use the $x$-intercepts to find where the modelled quantity ($y$) becomes zero, usually a natural boundary.
  4. State the domain as an inequality with its meaning: e.g. "$0 \leq w \leq 40$ where $w$ is the width in metres, width must be non-negative and at most 40 m."

Model limitations, what to say: NESA expects specific, contextual limitations. Good responses name an assumption the model makes and explain why that assumption might not hold.

  • "The model assumes the shape is exactly a parabola, but the projectile may be affected by wind resistance, changing the curve."
  • "The model predicts positive area for any width between 0 and 40 m, but practical constraints (e.g. minimum width for animals) would further restrict the domain."
  • "The profit model assumes a fixed price per unit, but in reality prices fall as supply increases."
Is $w = 50$ m valid for $A = 40w - w^2$? At $w = 50$: $A = 40(50) - 2500 = 2000 - 2500 = -500$ m². A negative area is impossible, $w = 50$ is outside the valid domain. The valid domain is $0 \leq w \leq 40$ (where area is non-negative).

A model's valid domain is the set of x-values that make physical sense in context, for example, time cannot be negative and a ball cannot be below the ground. Always state domain restrictions explicitly in HSC answers.

Pause, copy the valid domain definition (set of x-values that make physical sense) and the two common restriction types: x ≥ 0 for time/quantity, and an upper bound from the physical context into your book.

True or false: For the model $h = -4t^2 + 20t$ (height of a ball), the valid domain is $0 \leq t \leq 5$ because height can't be negative and the ball lands at $t = 5$.

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Solving Practical Problems from a Quadratic Graph
core concept

Once the valid domain is established, every exam question about a quadratic model asks for one of three things: the vertex (maximum height or minimum cost), an intercept (when a quantity starts or reaches zero), or a y-value at a specified x. Each technique involves reading the graph directly, always include units in your answer.

NESA provides you with a graph and asks you to answer questions about it, you must read values accurately, identify features, and translate numbers into contextual answers.

What NESA questions typically ask:

  • "What is the maximum [height/area/profit]?" → Read the $y$-coordinate of the vertex.
  • "When does [the ball land / the quantity return to zero]?" → Read the positive $x$-intercept.
  • "What is the [value/height/cost] at $t = k$?" → Read the $y$-value at the specified $x$.
  • "For what values of $x$ is [the quantity] greater than $k$?" → Find where the graph is above $y = k$ (read the two $x$-values where the graph meets the horizontal line $y = k$).
  • "State the valid values of $x$ for this model." → State the domain with a brief justification.
Reading "greater than" questions from a graph: Draw a horizontal line $y = k$. Read the two $x$-values where the parabola intersects this line. The parabola is above $y = k$ between these two values. State the answer as: "$x_a \leq x \leq x_b$, i.e. between $x_a$ and $x_b$ seconds/metres/units."
Mark scheme awareness: When answering "interpret the meaning of the vertex," NESA expects both the numerical answer AND the contextual meaning in one sentence. Giving only coordinates earns partial marks.

To solve practical problems from a quadratic graph: read the required feature directly (vertex for max/min, intercept for start/end, or a specific y-value by reading across then down). Always state units in your answer.

Pause, copy the three read-off techniques: read the vertex for max/min, read an intercept for start/end, read horizontally then vertically for a given x-value, and always state the unit of each answer.

Fill the blanks: For $A = 40w - w^2$, the maximum area is m² at a width of m. The valid domain is $0 \leq w \leq$ .

PROBLEM 1 · INTERPRETING FEATURES IN CONTEXT

A skateboard ramp is modelled by $h = -0.1x^2 + 2x$ where $h$ is height (m) and $x$ is horizontal distance (m). (a) What is the maximum height and where does it occur? (b) What is the horizontal length of the ramp? (c) State the valid domain.

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Part (a), maximum height (vertex)
$x$-intercepts: $h = 0 \Rightarrow -0.1x^2 + 2x = 0 \Rightarrow x(-0.1x + 2) = 0$
$x = 0$ or $x = 20$
Axis of symmetry: $x = (0 + 20)/2 = 10$ m
Find $x$-intercepts first to get the axis of symmetry.
PROBLEM 2 · PROFIT MODEL, READING FROM GRAPH

A company's weekly profit ($ thousands) is modelled by $P = -2n^2 + 20n - 32$ where $n$ is the number of items produced (hundreds). From the graph: (a) Find the break-even points (when $P = 0$). (b) Find the maximum profit and the production level that achieves it. (c) For what production levels does the company make a profit?

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Part (a), break-even ($P = 0$)
$-2n^2 + 20n - 32 = 0$
$n^2 - 10n + 16 = 0$
$(n - 2)(n - 8) = 0$
$n = 2$ or $n = 8$ (hundred items)
Break-even occurs at $n = 200$ items and $n = 800$ items, these are the $x$-intercepts of the profit graph.
PROBLEM 3 · LIMITATIONS

A quadratic model for a roller coaster height is $h = -0.02d^2 + 1.6d$ (m, over horizontal distance $d$ in metres). Identify two limitations of this model.

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Find the valid domain first
$x$-intercepts: $d = 0$ and $d = 80$ m
Valid domain: $0 \leq d \leq 80$ m
Outside this range the model gives negative heights, impossible. But knowing the domain helps identify where limitations bite.

Match each contextual question to the graph feature that answers it:

Top 3 list: For a profit model $P = -n^2 + 10n - 16$, list THREE things you must include in a complete answer to "fully analyse this quadratic model."

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Revisit your thinking

Farmer: $A = 40w - w^2$. $x$-intercepts at $w = 0$ and $w = 40$. Vertex: $w = 20$ m, $A = 400$ m². Maximum area is 400 m² at width 20 m. The valid domain is $0 \leq w \leq 40$, at $w = 50$, the model gives $A = -500$ m², which is physically impossible. The model breaks down outside the valid domain.

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Short answer, exam-style questions
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ApplyBand 44 marks

SA 1. A ball is thrown and its height is modelled by $h = -5t^2 + 25t$ (m, seconds). The graph is a downward parabola with $x$-intercepts at $t = 0$ and $t = 5$. (a) Interpret the $x$-intercepts in context. (b) Find the maximum height and when it occurs. (c) State the valid domain and justify your answer. (4 marks)

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ApplyBand 44 marks

SA 2. A company's monthly profit ($ thousands) is modelled by $P = -n^2 + 12n - 20$ where $n$ is items produced (hundreds). From the graph of this parabola: (a) Find the break-even points. (b) Find the maximum profit and the production level. (c) What production range gives a profit greater than $\$16{,}000$? (4 marks)

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EvaluateBand 54 marks

SA 3. A quadratic model for a ski jump is $h = -0.05d^2 + 3d$ where $h$ is height (m) and $d$ is horizontal distance (m). (a) Find the maximum height and horizontal distance at that point. (b) State the valid domain and explain why values outside this domain are invalid. (c) Give TWO limitations of this model for predicting the entire path of the ski jump. (4 marks)

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📖 Comprehensive answers (click to reveal)

SA 1 (4 marks): (a) $t=0$: ball is thrown from ground level (height 0 at start); $t=5$: ball returns to ground (lands 5 seconds after being thrown) [1]. (b) Axis: $t=(0+5)/2=2.5$ s; $h=-5(6.25)+25(2.5)=-31.25+62.5=31.25$ m. Max height 31.25 m at 2.5 s [1+1]. (c) $0\leq t\leq 5$, time can't be negative and the ball is on the ground (above ground) only between $t=0$ and $t=5$ [1].

SA 2 (4 marks): (a) $P=0$: $n^2-12n+20=0\Rightarrow(n-2)(n-10)=0\Rightarrow n=2$ or $n=10$ (hundreds). Break-even at 200 and 1000 items [1]. (b) Axis: $n=(2+10)/2=6$; $P=-36+72-20=16$. Max profit $\$16{,}000$ at 600 items [1+1]. (c) $P>16$: the vertex is the maximum, so $P=16$ only at the vertex, there is no range where $P>16$ (the parabola reaches exactly 16 at the vertex). If the question intended $P>15$: solve $-n^2+12n-20>15\Rightarrow n^2-12n+35<0\Rightarrow (n-5)(n-7)<0\Rightarrow 5<n<7$, i.e. 500 to 700 items [1].

SA 3 (4 marks): (a) $h=0$: $d(3-0.05d)=0\Rightarrow d=0$ or $d=60$. Axis: $d=30$ m; $h=-0.05(900)+3(30)=-45+90=45$ m. Max height 45 m at 30 m horizontal distance [1]. (b) Domain $0\leq d\leq 60$; outside this, the model gives $h<0$ (below ground), which is physically impossible once the jump ends [1]. (c) Any TWO: model assumes single symmetric parabola, but real jump includes takeoff ramp at an angle (not zero); landing may involve different dynamics; air resistance and lift from skis alter the path; the skier's body position changes the aerodynamic profile [1 per valid limitation].

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Boss battle · The Quadratic Analyst
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Science Jump · platform challenge

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