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Year 12 Maths Advanced (2027) Functions

Modelling with Functions & Transformations

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Further graph transformations and modelling

Modelling with functions and transformations chooses a suitable function — linear, quadratic, exponential or trigonometric — and transforms it with \(y=af\!\big(b(x+c)\big)+d\) to match data and solve practical problems.

Part of the NSW Year 12 Mathematics Advanced course, in the Functions area of study (Further graph transformations and modelling focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

Modelling with functions uses a transformed standard function to describe a real situation. This Year 12 Mathematics Advanced topic (MAV-12-02) uses quadratic models in vertex form \(h=a(t-p)^2+q\) for a maximum or minimum, and exponential models \(P=Ab^{\,t}\) for growth and decay, without calculus.

A quadratic model in vertex form \(h=a(t-p)^2+q\) has its turning point at \((p,q)\): a maximum if \(a<0\), a minimum if \(a>0\), with value \(q\) at \(t=p\).

An exponential model \(P=Ab^{\,t}\) starts at \(A\) when \(t=0\); it grows if \(b>1\) and decays if \(0

Projectile parabolah = -5(t-2)^2 + 20 has its vertex at (2,20), the maximum height 20 m at t = 2 s. th (2, 20) 2 4
Projectile \(h=-5(t-2)^2+20\): maximum \(20\) m at \(t=2\) s.
Exponential growth modelAn exponential model grows faster and faster over time. tP P = A b^t (b>1)
An exponential model \(P=Ab^{\,t}\) with \(b>1\) grows ever faster.
\[h=a(t-p)^2+q\quad\text{vertex }(p,q)\]
\[P=A\,b^{\,t}\quad(\text{start }A;\ b>1\text{ grows},\ 0
quadratic vertex form h equals a times (t minus p) squared plus q; exponential P equals A times b to the t

Method

  1. Choose the model: quadratic for a turning point, exponential for a constant percentage change.
  2. Read or substitute: vertex \((p,q)\); value at a time by substituting \(t\).
  3. Solve for a time by setting the model equal to the target value.
Example 1 — Maximum
\(h=-5(t-2)^2+20\) m. Find the maximum height and when.
Solution
\(\text{vertex}\)\(=\)\((2,20)\)
\(\text{max}\)\(=\)\(20\text{ m at }t=2\text{ s}\)
Example 2 — When zero
For \(h=-5(t-2)^2+20\), when does \(h=0\)?
Solution
\(5(t-2)^2\)\(=\)\(20\)
\((t-2)^2\)\(=\)\(4\)
\(t\)\(=\)\(4\ (t\ge0)\)
Example 3 — Exponential
\(P=200(1.1)^t\). Find \(P\) at \(t=3\).
Solution
\(P\)\(=\)\(200(1.1)^3\)
\(=\)\(200\times1.331\approx266\)
Example 4 — Growth model
\(P=50\times2^{\,t}\): initial number, value at \(t=3\), and when \(P=800\)?
Solution
\(P(0)\)\(=\)\(50\)
\(P(3)\)\(=\)\(50\times8=400\)
\(2^{\,t}\)\(=\)\(16\Rightarrow t=4\)

Common pitfalls

Vertex form reads off directly. \(a(t-p)^2+q\) turns at \((p,q)\) — note the minus in \((t-p)\).
Sign of \(a\). \(a<0\) opens down (a maximum); \(a>0\) opens up (a minimum).
Start value is \(A\). In \(P=Ab^{\,t}\) the value at \(t=0\) is \(A\), not \(b\).

Frequently asked questions

How do you find the maximum of a quadratic model?

Write it in vertex form h = a(t - p) squared + q. The turning point is at (p, q), so if a is negative the maximum value is q, reached at t = p.

What does the vertex of a parabola tell you in a model?

The vertex gives the best or worst value of the model, such as the greatest height of a projectile or the lowest cost, and the time at which it occurs.

How do you find the starting value of an exponential model?

Substitute t = 0. In P = A times b to the t, the starting value is A because b to the power 0 is 1.

How do you tell if an exponential model grows or decays?

It grows when the base b is greater than 1 and decays when b is between 0 and 1.