Constructing Exponential Models from Initial Values and Percentage Changes

Hello! In the previous lesson, you found the maximum or minimum of a quadratic model and checked whether its input made sense in context. This lesson asks a different modelling question: what if a quantity changes by the same percentage each period? You will turn an initial value and a percentage increase or decrease into a function, then check that the function means what you intend.

The difference between adding and multiplying

Imagine two groups of stores, each starting with 100 stores. Group A adds 50 stores every year. Group B grows by 50% every year. They have the same starting value and even the same total after one year, but their rules differ.

Number of stores against years: the blue line represents 50 additional stores per year; the orange curve represents growth of 50% of the current number per year. Both start at 100.

For Group A, “50 more” is a fixed amount. For Group B, “50% more” depends on how many stores it already has. Its first increase is 50 stores, but its next is 50% of 150, or 75 stores.

Years elapsed, $x$	A: add 50 each year	B: multiply by $1.5$ each year
$0$	$100$	$100$
$1$	$150$	$150$
$2$	$200$	$225$
$3$	$250$	$337.5$

The rules behind the graph are

A (x) = 100 + 50 x and B (x) = 100 (1.5)^{x} .

The linear model has a constant difference of 50 for each additional year. The exponential model has a constant multiplier of $1.5$ : each year’s value is 150% of the previous year’s value. The exponent counts how many yearly multiplications have occurred.

A model can produce a fractional number of stores, such as $337.5$ . That is a prediction from a simplified rule, not a claim that half a store exists; an actual store count would be a whole number.

Constructing a growth model

Suppose an account starts with 800 currency units and increases by 4% at the end of each year, with no deposits or withdrawals. Let $S (n)$ be its value after $n$ years. At the start, $S (0) = 800$ .

A 4% increase means retaining the original 100% and adding another 4%. In decimal form, the new value is 104% of the old one, so the yearly growth factor is $1.04$ :

S (1) = 800 (1.04) = 832.

In the second year, the 4% applies to 832, not to the original 800. Thus

S (2) = 800 (1.04) (1.04) = 800 (1.04)^{2} = 865.28.

The same multiplication repeats for every year, giving

S (n) = 800 (1.04)^{n} .

Notice the two jobs in this formula. The initial value, 800, tells you the output at $n = 0$ , because $(1.04)^{0} = 1$ . The factor, $1.04$ , tells you what happens during one year. The exponent tells you how many years of change have taken place.

In general, if a quantity starts at $a$ and increases by $p$ each period, where $p$ is the percentage written as a decimal, its model is

F (n) = a (1 + p)^{n} .

For example, a 4% increase uses $p = 0.04$ , not $p = 4$ . Because $1 + p > 1$ , repeated multiplication produces growth. The value $1.04$ is the factor; $0.04$ , or 4%, is the percentage rate. They are related, but they are not interchangeable.

Constructing a decay model

Now suppose a piece of equipment is initially worth 600 currency units and loses 12% of its value each year. A loss of 12% means it keeps 88% of its previous value. Its yearly factor is therefore $0.88$ , not $0.12$ .

If $V (n)$ is its modelled value after $n$ years, then

V (n) = 600 (0.88)^{n} .

The model gives $V (0) = 600$ , $V (1) = 528$ , and $V (2) = 464.64$ . During the first year the modelled loss is 72 currency units; during the second it is 63.36. The percentage lost stays at 12%, but the amount lost changes because the percentage is taken from a smaller value.

For a decrease of $p$ each period, the general form is

F (n) = a (1 - p)^{n},

where $p$ is again a positive percentage written as a decimal. For an ordinary percentage decrease smaller than 100%, the factor lies between 0 and 1. Repeated multiplication by such a factor makes the output smaller while keeping it positive.

Watch how both types of model are assembled from a starting amount and a percentage. In the video, pay particular attention to the distinction between the stated percentage and the multiplier placed inside the formula.

Exponential Functions with Percent Increase or Decrease

In “Exponential Functions with Percent Increase or Decrease,” Tom Teaches Math works through one increase and one decrease. Watch to reinforce why the factor must include the part of the quantity that remains.

Watch the increase, beginning with the 15% annual-interest example; focus on why its factor is 1.15, rather than 15 or 0.15. Then watch the decrease, beginning with the vehicle-value example; focus on why an 8% decrease leaves a factor of 0.92. The narrator sometimes calls the multiplier a “growth rate”; in your own working, distinguish the percentage rate from the resulting growth or decay factor.

Match the exponent to the stated period

So far, each percentage has applied once per year, and the exponent has counted years. But the length of one change period is what matters—not the word “year” itself.

Suppose equipment initially worth 900 currency units loses 20% of its value every two years. Let $E (t)$ represent its value $t$ years from now. It retains 80% of its value per two-year period, so the factor is $0.8$ . The number of completed two-year periods after $t$ years is $t /2$ . A model at two-year intervals is

E (t) = 900 (0.8)^{t /2}, t = 0, 2, 4, 6, \dots

Check the timing: at $t = 2$ , the exponent is 1 and the model applies the decrease once. At $t = 4$ , the exponent is 2, so

E (4) = 900 (0.8)^{2} = 576.

Using $900 (0.8)^{t}$ instead would apply a 20% decrease every year, which is not the rule supplied. The expression with $t /2$ can also be used between the two-year marks if we choose to extend it as a smooth model; the stated rule alone directly specifies the values at completed two-year intervals.

Check what your function claims

When you construct an exponential model, three short checks catch many errors:

Initial value: Substituting zero periods should return the stated starting amount.
One period: Substituting one change period should give the stated percentage increase or decrease.
Context: State what the input measures, what the output measures, and which inputs make sense.

For the annual account model, $n = 0, 1, 2, \dots$ describes values just after whole numbers of years when the increase is applied at each year’s end. A smooth exponential curve may be useful in other contexts, but it assumes the same pattern can be extended between those recorded times. Likewise, a constant percentage is a modelling assumption, not a promise that a real account, population, or item of equipment will keep changing at that rate indefinitely.

The central distinction is straightforward: a constant amount suggests addition; a constant percentage suggests multiplication. Once you have identified the initial value, the factor, and the length of one period, the exponential function follows.

Conclusion

To construct an exponential model, write the initial value first. Convert the percentage to a decimal, then use $1 + p$ for an increase or $1 - p$ for a decrease. Raise that factor to the number of change periods, and check the model at zero and after one period.

Next, you will estimate an instantaneous rate of change using a tangent. Even when an exponential model has a constant percentage change, its absolute change—and therefore the steepness of its graph—does not stay constant.