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Year 11 Methods (Unit 1 & 2) Functions, Relations And Transformations

Functions And Modelling Exercises

20 practice questions 1 video lesson Theory + worked examples

Apply function models for Queensland Year 11 Mathematical Methods (QCAA). This topic brings functions to life, using function notation to describe real situations and answer practical questions without any calculus.

You will learn to evaluate and build a rule, solve for an input, state a domain and range, use piecewise functions, and find a maximum or minimum from the turning point of a quadratic — modelling skills prized across QCAA assessment.

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Theory

In Year 11 Mathematical Methods (QCAA), a function model uses function notation \(f(x)\) to describe a real situation. This page shows how to evaluate a model, build a rule, solve \(f(x)=\text{value}\), state a sensible domain and range, and find a maximum or minimum from the vertex of a quadratic — all without calculus.

Function notation \(f(x)\) names a rule: \(f(a)\) is the output when the input is \(a\). The independent variable is the input (often time or length) and the dependent variable is the output (a cost, height or volume).

The domain is the set of allowed inputs and the range is the set of resulting outputs; in a model these are limited by what makes physical sense (you cannot have a negative time or length). A piece-wise function uses different rules on different parts of the domain, such as a flat fee for a short trip and a per-kilometre rate after that.

Model, then interpret. Write the rule, do the algebra, then read the answer back in the words and units of the problem (dollars, metres, minutes).
A linear cost modelStraight line C=60 plus 90 t; a fixed cost of 60 dollars plus 90 dollars per hour. x y (0,60)
A linear model \(C=60+90t\): a fixed \(\$60\) plus \(\$90\) per hour.
A quadratic model with a maximumParabola A=x times (20 minus x) with its greatest value at the vertex ten, one hundred. x y (10,100)
A quadratic model \(A=x(20-x)\): the greatest value is at the vertex \((10,\,100)\).

Evaluate a model at an input \(a\):

\[y=f(a)\]
y=f(a)

Maximum or minimum of a quadratic \(y=at^2+bt+c\) is at the axis of symmetry:

\[t=-\dfrac{b}{2a}\]
t=-b2a
Which is it? A quadratic model opens down when \(a\lt 0\) (a maximum) and up when \(a\gt 0\) (a minimum); the vertex gives the greatest or least value.

How to solve a modelling problem

  1. Set up: name the variables and write the rule \(f(x)\) from the description.
  2. Do the maths: evaluate \(f(a)\), solve \(f(x)=\text{value}\), or find the vertex \(t=-\dfrac{b}{2a}\) for a maximum or minimum.
  3. Interpret: state the answer with correct units and a sensible domain or range for the context.
Example 1 — Evaluate a model
A plumber charges a \(\$60\) call-out fee plus \(\$90\) per hour, so the cost of a \(t\)-hour job is \(C(t)=60+90t\). Find the cost of a \(4\)-hour job.
Solution

Substitute \(t=4\) into the model:

\(C(4)\)\(=\)\(60+90(4)\)
\(=\)\(60+360\)
\(=\)\(420\)

The cost is \(\$420\).

Plumber cost modelLine C=60 plus 90 t; a four hour job costs 420 dollars. x y (4,420)
C(4)=420
Example 2 — Solve f(x) = value
A tank drains so that its volume after \(t\) minutes is \(V(t)=800-25t\) litres. When does it hold \(300\) litres, and when is it empty?
Solution

Set \(V(t)=300\):

\(800-25t\)\(=\)\(300\)
\(-25t\)\(=\)\(300-800\)
\(-25t\)\(=\)\(-500\)
\(t\)\(=\)\(20\)

Empty means \(V(t)=0\):

\(800-25t\)\(=\)\(0\)
\(25t\)\(=\)\(800\)
\(t\)\(=\)\(32\)

It holds \(300\) L at \(t=20\) min and is empty at \(t=32\) min.

Draining tank modelLine V=800 minus 25 t; the tank holds 300 litres at 20 minutes and is empty at 32 minutes. x y (20,300) (32,0)
t=20,t=32
Example 3 — Maximum from the vertex
A ball's height is \(h(t)=-5t^2+20t+15\) metres, \(t\) seconds after release. Find the greatest height and when it occurs.
Solution

Axis of symmetry with \(a=-5,\ b=20\):

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

Greatest height — substitute \(t=2\):

\(h(2)\)\(=\)\(-5(2)^2+20(2)+15\)
\(=\)\(-20+40+15\)
\(=\)\(35\)

The greatest height is \(35\) m, reached at \(t=2\) s.

Projectile height modelParabola h=minus 5 t squared plus 20 t plus 15 reaching its greatest height 35 metres at 2 seconds. x y (2,35)
h=35
Example 4 — Build and use a piece-wise model
A taxi charges a flat \(\$5\) for the first \(2\) km, then \(\$2\) per km after that. Write the fare \(C(d)\) as a piece-wise rule and find the cost of a \(5\) km trip.
Solution

Build the rule — flat, then \(\$2\) per extra km beyond \(2\):

\(C(d)\)\(=\)\(5,\quad 0\lt d\le 2\)
\(=\)\(5+2(d-2),\quad d\gt 2\)

For a \(5\) km trip, \(d=5\gt 2\), so use the second piece.

Substitute \(d=5\):

\(C(5)\)\(=\)\(5+2(5-2)\)
\(=\)\(5+2(3)\)
\(=\)\(5+6\)
\(=\)\(11\)

The fare is \(\$11\) for the \(5\) km trip.

Piece-wise taxi fare modelA flat 5 dollars up to 2 km, then rising at 2 dollars per km; a 5 km trip costs 11 dollars. x y (5,11)
C(5)=11

Common pitfalls

Ignoring the domain. A model such as \(V(t)=800-25t\) only makes sense for \(0\le t\le 32\); a negative time or a negative volume has no meaning.
Reading the vertex \(t\)-value as the answer. \(t=-\dfrac{b}{2a}\) tells you when the maximum occurs; substitute it back to get the maximum value.
Using the wrong piece. In a piece-wise model, check which interval the input falls in before you substitute.

Frequently asked questions

What does f(a) mean in a model?

It is the output of the model when the input is \(a\); substitute \(x=a\) into the rule, for example \(C(4)=60+90(4)=420\).

How do I find a maximum without calculus?

For a quadratic \(y=at^2+bt+c\), the maximum or minimum is at the axis of symmetry \(t=-\dfrac{b}{2a}\); substitute that \(t\) back to get the value.

How do I state a sensible domain for a model?

Restrict the input to values that make physical sense — for instance \(0\le t\le 32\) for a tank that empties at \(t=32\) minutes.

What is a piece-wise function?

A function that uses different rules on different parts of the domain, such as a flat fee for a short trip and a per-kilometre rate beyond a set distance.

How do I know if a quadratic model has a maximum or a minimum?

Look at the coefficient of \(t^2\): if it is negative the parabola opens down (a maximum); if positive it opens up (a minimum).