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Year 12 Maths Standard 2 (2027) Algebraic relationships

Exponential Models (Growth & Decay)

20 practice questions 0 video lessons Theory + worked examples

Master exponential models of growth and decay for NSW Year 12 Mathematics Standard 2. In this topic you construct and analyse models of the form \(y=k\,a^{x}\) and \(y=k\,a^{-x}\), reading the initial value \(k\) off the \(y\)-intercept and using the base \(a\) as the growth or decay factor.

You will learn to decide growth \((a>1)\) from decay \((0<a<1)\), find the percentage change each period, evaluate a model at a given value, and explain the limitations of using an exponential model to predict real quantities such as populations, savings, medicine and depreciation far into the future β€” a core Standard 2 skill.

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Theory

Exponential models \(y=k\,a^{x}\) and \(y=k\,a^{-x}\) describe quantities that change by a fixed factor each step. This Year 12 Standard 2 (NSW) guide shows how to read the initial value \(k\), use the growth or decay factor \(a\), find the percentage change per period, and explain the limitations of an exponential model.

An exponential model has the form \(y=k\,a^{x}\) (or \(y=k\,a^{-x}\)), where the variable sits in the exponent. It describes a quantity that is multiplied by the same factor \(a\) every time \(x\) increases by 1.

The number \(k\) is the initial value: because \(a^{0}=1\), putting \(x=0\) gives \(y=k\), so \(k\) is the \(y\)-intercept of the graph. The base \(a\) is the growth or decay factor.

If \(a>1\) the amount rises each step β€” exponential growth. If \(0<a<1\) the amount falls each step β€” exponential decay. A negative index form \(y=k\,a^{-x}\) equals \(k\left(\tfrac{1}{a}\right)^{x}\), so it also models decay. This is a core NSW Year 12 Mathematics Standard 2 skill.

Exponential growth curveA growth curve y=k a^x with base a greater than 1, starting at k=200 and rising steeply. t B 0 200 2 800
Growth \((a>1)\): starts at \(k\) and rises faster and faster.
Exponential decay curveA decay curve y=k a^x with a between 0 and 1, starting at k=1000 and falling towards zero. t P 0 1000
Decay \((0<a<1)\): starts at \(k\) and falls towards zero.

The two exponential model forms are:

\[y = k\,a^{x} \qquad y = k\,a^{-x}\]
y=kax

The initial value is read at \(x=0\), using \(a^{0}=1\):

\[y(0) = k\,a^{0} = k\]
y(0)=k

The percentage change per step comes straight from the base \(a\):

\[\text{growth }=(a-1)\times100\%\qquad\text{decay }=(1-a)\times100\%\]
(a-1)×100%
Growth vs decay. The base decides it: \(a>1\) grows, \(0<a<1\) decays. The size of \(k\) never changes which one it is.

How to build and read an exponential model

  1. Find \(k\) β€” the starting amount is the initial value and the \(y\)-intercept.
  2. Find \(a\) β€” a \(p\%\) rise gives \(a=1+\tfrac{p}{100}\); a \(p\%\) fall gives \(a=1-\tfrac{p}{100}\). From a table, divide consecutive \(y\)-values.
  3. Write the model \(y=k\,a^{x}\) (or \(y=k\,a^{-x}\)).
  4. Evaluate by working out the power first, then multiplying by \(k\).
  5. Analyse β€” decide growth or decay, interpret the intercept, and remember the model has limits far into the future.
Example 1 β€” Growth: a savings account
A savings account is modelled by \(A=4000\times(1.05)^{t}\) dollars after \(t\) years. Find the initial balance, the balance after \(10\) years, and the yearly growth rate.
Solution

The initial balance is the \(y\)-intercept \(k\); evaluate the power for \(t=10\).

Savings growth curveBalance A=4000*1.05^t growing from $4000 to about $6516 over 10 years. t A 0 4000 10 6516
\(A(0)\)\(=\)\(4000\times(1.05)^{0}=\$4000\)
\(A(10)\)\(=\)\(4000\times(1.05)^{10}\)
\(\)\(=\)\(4000\times 1.6289\)
\(\)\(=\)\(\$6515.58\)
\(\text{rate}\)\(=\)\(1.05-1=5\%\)
A=6515.58
Example 2 β€” Decay: a drug in the blood
A drug clears from the bloodstream as \(D=60\times(0.5)^{t}\) mg after \(t\) hours. Find the initial dose, the amount left after \(3\) hours, and the percentage lost each hour.
Solution

Substitute \(t=0\) then \(t=3\); the percentage lost is \(1-a\).

Drug decay curveDrug amount D=60*0.5^t halving each hour, from 60 mg to 7.5 mg after 3 hours. t D 0 60 3 7.5
\(D(0)\)\(=\)\(60\times(0.5)^{0}=60\text{ mg}\)
\(D(3)\)\(=\)\(60\times(0.5)^{3}\)
\(\)\(=\)\(60\times 0.125\)
\(\)\(=\)\(7.5\text{ mg}\)
\(\text{lost}\)\(=\)\(1-0.5=50\%\)
D=7.5
Example 3 β€” Negative index form
A quantity is modelled by \(y=500\times 3^{-x}\). Find \(y\) when \(x=2\), and state whether the model shows growth or decay.
Solution

A negative index means a reciprocal power; simplify \(3^{-2}\) first.

Negative-index decay curveCurve y=500*3^(-x) falling from 500 to about 55.6 at x=2. x y 0 500 2 55.6
\(y\)\(=\)\(500\times 3^{-2}\)
\(3^{-2}\)\(=\)\(\dfrac{1}{3^{2}}=\dfrac{1}{9}\)
\(y\)\(=\)\(500\times\dfrac{1}{9}\)
\(y\)\(\approx\)\(55.56\)

Since \(3^{-x}=\left(\tfrac{1}{3}\right)^{x}\) and \(\tfrac{1}{3}<1\), this is decay.

Example 4 β€” Construct and analyse
A new cafe starts with \(150\) customers a week and numbers grow by \(8\%\) each week. Write a model, predict the weekly customers after \(12\) weeks, and give one limitation.
Solution

A \(8\%\) rise means \(a=1.08\); build \(N=k\,a^{t}\) then evaluate.

Cafe customer growth curveWeekly customers N=150*1.08^t rising from 150 to about 378 over 12 weeks. t N 0 150 12 378
\(a\)\(=\)\(1+0.08=1.08\)
\(N\)\(=\)\(150\times(1.08)^{t}\)
\(N(12)\)\(=\)\(150\times 2.5182\)
\(N(12)\)\(\approx\)\(378\text{ customers}\)

Limitation: a fixed \(8\%\) growth cannot continue forever β€” seating and the local market will slow it, so the model overestimates far ahead.

Common pitfalls

The base decides growth or decay. Not the size of \(k\). A big starting value with \(0<a<1\) still decays towards zero.
The base is a multiplier, not a percentage. \(a=1.06\) is \(+6\%\) per step and \(a=0.85\) is \(-15\%\) β€” do not read \(0.85\) as \(85\%\) lost.
Powers before products. Work out \(a^{x}\) first, then multiply by \(k\); and remember no exponential model grows without limit forever.

Frequently asked questions

What is an exponential model?

It is a relationship of the form y equals k times a to the power x, where the variable is in the exponent. The quantity is multiplied by the same factor a each time x increases by one, which makes it grow or decay much faster than a straight line.

What do k and a mean in y equals k a to the x?

k is the initial value, the amount when x is zero, and it is the y-intercept of the graph because a to the power zero is one. a is the growth or decay factor, the number you multiply by each step.

How do you tell growth from decay?

Look only at the base a. If a is greater than 1 the amount rises each step, which is exponential growth. If a is between 0 and 1 the amount falls each step, which is exponential decay. The size of k does not change this.

How do you find the percentage change per period?

For growth the increase is a minus 1 as a percentage, so a equals 1.05 means 5 percent growth. For decay the decrease is 1 minus a as a percentage, so a equals 0.8 means 20 percent decay each period.

What does y equals k a to the minus x mean?

A negative index is a reciprocal power, so y equals k a to the minus x is the same as k times one over a all to the power x. Because one over a is less than 1 when a is greater than 1, this form models decay.

What is a limitation of an exponential model?

A constant percentage growth cannot continue forever. Real limits such as space, food, resources or market size eventually slow the growth, so an exponential model tends to overestimate a quantity far into the future.