Year 11 Methods (Unit 1 & 2) Quadratics

Quadratic Models

Q: What is a quadratic model?

A parabola, y equals a x squared plus b x plus c, used to describe a real situation such as a projectile's height, an area, or a profit.

Q: How do I find the maximum or minimum of a quadratic model?

Find the input x equals minus b over two a at the turning point, then substitute it back into the model to get the maximum or minimum value.

Q: How do I find when a model reaches the ground or breaks even?

Set the model equal to zero and solve the quadratic. Those zeros are the ground time or the break-even inputs.

Q: Is the model a maximum or a minimum?

It is a maximum when a is less than zero, so the parabola opens downwards, and a minimum when a is greater than zero, so it opens upwards.

Q: What does the y-intercept mean in a model?

It is the value at input zero, which is the starting height, initial amount, or fixed cost depending on the context.

Q: Do I need calculus to find the turning point?

No. In Year 11 you use the axis of symmetry, x equals minus b over two a, not differentiation.

20 practice questions 1 video lesson Theory + worked examples

Learn to use quadratic models for Queensland Year 11 Mathematical Methods (QCAA). A quadratic model uses a parabola to describe a real situation, such as a projectile's height or a profit.

You will learn to evaluate a model, find its zeros for ground or break-even values, and locate the maximum or minimum using the turning point — the applied heart of the topic, connecting parabolas to motion, area and money problems.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 1 video(s)

Quadratic Models - Video - Modeling with Quadratic Functions Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), a quadratic model uses a parabola \(y=ax^2+bx+c\) to describe a real situation — a thrown ball's height, an enclosed area, a business's profit. This page shows how to evaluate a model, find its zeros (ground or break-even values), and find the maximum or minimum using the turning point at \(x=-\dfrac{b}{2a}\).

A quadratic model is a parabola used to represent a quantity that rises then falls (or falls then rises), such as the height of a projectile over time or the area of an enclosure against its width.

The zeros (where the model equals \(0\)) mark meaningful moments — when a ball hits the ground, or when profit is zero (break-even). The y-intercept is the starting value, at input \(0\).

The turning point (vertex) gives the maximum value when the parabola opens downwards (\(a<0\)) or the minimum when it opens upwards (\(a>0\)). It sits on the axis of symmetry \(x=-\dfrac{b}{2a}\).

Max/min is a two-step read. The input at which it occurs is \(x=-\dfrac{b}{2a}\); the max/min value is what you get by substituting that \(x\) back into the model.

A height-versus-time model: starts at the \(y\)-intercept, peaks at the vertex, and reaches the ground at a zero.

The maximum value is the height of the turning point; the zeros are where the model meets the axis.

The input at the turning point (axis of symmetry) of \(y=ax^2+bx+c\):

\[x=-\dfrac{b}{2a}\]

x = - \frac{b}{2 a}

The maximum or minimum value is the model evaluated there:

\[y_{\text{max/min}}=f\!\left(-\dfrac{b}{2a}\right)\]

y_{\max / \min} = f (- \frac{b}{2 a})

Which way does it open? \(a<0\) opens downwards so the vertex is a maximum; \(a>0\) opens upwards so it is a minimum.

How to work with a quadratic model

Read the question: decide whether it asks for a value at an input (evaluate), a zero (set the model to \(0\)), or a max/min (turning point).
Apply the right tool: substitute the input, or solve the quadratic for zeros, or find \(x=-\dfrac{b}{2a}\) for the turning point.
Interpret in context: attach units, reject any answer that makes no sense (such as a negative time), and state what the value means.

Example 1 — Evaluate the model

A ball is thrown so its height is \(h=-5t^2+20t+25\) (metres, \(t\) seconds). Find the initial height and the height after \(2\) seconds.

Solution

Initial height — substitute \(t=0\):

\(h\)	\(=\)	\(-5(0)^2+20(0)+25\)
	\(=\)	\(25\)

Height at \(t=2\) — substitute \(t=2\):

\(h\)	\(=\)	\(-5(2)^2+20(2)+25\)
	\(=\)	\(-20+40+25\)
	\(=\)	\(45\)

Initial height \(25\) m; after \(2\) s the ball is at \(45\) m.

h (0) = 25, h (2) = 45

Example 2 — Find a zero (when it lands)

For the same ball, \(h=-5t^2+20t+25\). When does it hit the ground?

Solution

The ground is \(h=0\):

\(-5t^2+20t+25\)

\(=\)

\(0\)

Divide by \(-5\), then factorise:

\(t^2-4t-5\)	\(=\)	\(0\)
\((t-5)(t+1)\)	\(=\)	\(0\)
\(t\)	\(=\)	\(5 \ \text{or}\ -1\)

Time cannot be negative, so reject \(t=-1\).

The ball hits the ground after \(5\) seconds.

t = 5

Example 3 — Find the maximum value

For \(h=-5t^2+20t+25\), find the maximum height and when it occurs.

Solution

Turning point input — \(t=-\dfrac{b}{2a}\) with \(a=-5,\ b=20\):

\(t\)	\(=\)	\(-\dfrac{20}{2(-5)}\)
	\(=\)	\(-\dfrac{20}{-10}\)
	\(=\)	\(2\)

Maximum height — substitute \(t=2\):

\(h\)	\(=\)	\(-5(2)^2+20(2)+25\)
	\(=\)	\(-20+40+25\)
	\(=\)	\(45\)

Since \(a=-5<0\), the parabola opens downwards, so this is a maximum.

Maximum height \(45\) m, reached at \(t=2\) s.

h_{\max} = 45

Example 4 — Maximise an area

A farmer fences a rectangular yard against a wall using \(60\) m of fencing for the three open sides. If the width is \(x\) m, the area is \(A=x(60-2x)\). Find the width that maximises the area, and that area.

Solution

Expand to standard form:

\(A\)	\(=\)	\(x(60-2x)\)
	\(=\)	\(-2x^2+60x\)

Turning point — \(x=-\dfrac{b}{2a}\) with \(a=-2,\ b=60\):

\(x\)	\(=\)	\(-\dfrac{60}{2(-2)}\)
	\(=\)	\(-\dfrac{60}{-4}\)
	\(=\)	\(15\)

Maximum area — substitute \(x=15\):

\(A\)	\(=\)	\(15(60-2(15))\)
	\(=\)	\(15(30)\)
	\(=\)	\(450\)

Width \(15\) m gives the maximum area of \(450\) m\(^2\).

A = 450

Common pitfalls

Reporting \(x=-\dfrac{b}{2a}\) as the maximum. That is the input where the max/min occurs; the max/min value comes from substituting it back into the model.

Keeping an impossible answer. Reject negative times, negative lengths and other values that make no sense in the context.

Missing the initial value. The starting amount is the \(y\)-intercept — substitute input \(0\), not the vertex.

Sign of \(a\) and max versus min. \(a<0\) gives a maximum, \(a>0\) gives a minimum; check before you state which one you found.

Frequently asked questions

What is a quadratic model?

A parabola \(y=ax^2+bx+c\) used to describe a real situation, such as a projectile's height, an area, or a profit.

How do I find the maximum or minimum of a quadratic model?

Find the input \(x=-\dfrac{b}{2a}\) at the turning point, then substitute it back into the model to get the maximum or minimum value.

How do I find when a model reaches the ground or breaks even?

Set the model equal to \(0\) and solve the quadratic; those zeros are the ground time or break-even inputs.

Is the model a maximum or a minimum?

It is a maximum when \(a<0\) (opens downwards) and a minimum when \(a>0\) (opens upwards).

What does the y-intercept mean in a model?

It is the value at input \(0\) — the starting height, initial amount, or fixed cost, depending on the context.

Do I need calculus to find the turning point?

No — in Year 11 you use the axis of symmetry \(x=-\dfrac{b}{2a}\), not differentiation.

← Previous subtopic

Families Of Quadratic Polynomial Functions

Next subtopic →

This is the last subtopic

Quadratic Models

📖 Theory

How to work with a quadratic model

Common pitfalls

Frequently asked questions

Theory