Maths · 1 Number

1.16 Exponential growth and decay

Before you start

Percentage multipliers: $\times 1.15$ for a 15% increase and $\times 0.85$ for a 15% decrease (subtopic 1.12).
Compound interest as repeated multiplication, e.g. $1200 \times {1.025}^{4}$ .
Index notation, including $k^{0} = 1$ , and the power and root keys on your calculator.
Square roots and cube roots, e.g. $\sqrt{1.21} = 1.1$ and $\sqrt[3]{0.729} = 0.9$ .
Drawing graphs and finding where two graphs meet on a graphic display calculator (subtopic 3.2).

Syllabus objectives covered (1)

Use exponential growth and decay.

Part 1 of 6

Multiplying by the same number each time

Some quantities change by the same amount each period, for example a jar that gets $100 more every year. That is linear change. Other quantities change by the same percentage each period. That is exponential change, and it is what this subtopic is about.
Take $1000 that grows by 10% each year. Each year you multiply by $1.1$ , the multiplier for a 10% increase from the percentages topic.
$1000 \to 1100 \to 1210 \to 1331$
Look at the increases: $100, then $110, then $121. They get bigger each year, because each 10% is 10% of a bigger amount. This is exponential growth.
Now take $1000 that falls by 10% each year. The multiplier for a 10% decrease is $0.9$ .
$1000 \to 900 \to 810 \to 729$
The decreases are $100, $90, $81. They get smaller, because each 10% is taken from a smaller amount. This is exponential decay.
So the multiplier tells you which one you have. A multiplier greater than $1$ gives growth. A multiplier between $0$ and $1$ gives decay.
Depreciation is a fall in the value of an item, such as a car or a machine, over time. 'Depreciates by 12% per year' means that each year the item loses 12% of its value at the start of that year. So you multiply by $0.88$ each year.
Population change works the same way. 'The population grows by 2.5% per year' means multiply by $1.025$ each year. 'The population falls by 4% per year' means multiply by $0.96$ each year.
You have already met one case: compound interest is exponential growth of money. Everything in this lesson uses the same method.

Part 2 of 6

The formula, and counting the periods

Call the start value $V_{0}$ and the multiplier $k$ . After one period the value is $V_{0} \times k$ . After two periods it is $V_{0} \times k \times k = V_{0} \times k^{2}$ . Each period puts one more factor of $k$ on the end.
So after $n$ periods the value is:
$V = V_{0} \times k^{n}$
When the change is $r %$ per period, the multiplier is $1 + \frac{r}{100}$ for growth and $1 - \frac{r}{100}$ for decay:
$growth: V = V_{0} {(1 + \frac{r}{100})}^{n} decay: V = V_{0} {(1 - \frac{r}{100})}^{n}$
These are not new facts to learn. They come straight from the multiplier, so if you can build the multiplier from the percentage, you can build the formula.
Example: a car costs $18 500 and depreciates by 12% per year. Its value after 3 years is $18 500 \times {0.88}^{3} = 12 607.232$ , which is $12 607.23 to the nearest cent.
The power $n$ is the number of periods, not a date or a clock time. From 2020 to 2026 is $2026 - 2020 = 6$ years, so $n = 6$ . If the period is 20 minutes, then 2 hours is $120 \div 20 = 6$ periods, so $n = 6$ again.
When $n = 0$ the formula gives $V_{0} \times k^{0} = V_{0}$ , because $k^{0} = 1$ . So in a formula such as $P = 24 000 \times {1.03}^{t}$ , the number in front of the power, $24 000$ , is the value at the start.
Write the calculation on the page with its numbers, for example $18 500 \times {0.88}^{3}$ . That is the evidence of your method. Then enter it in one go and round only the final answer: to 3 significant figures, unless the question asks for the nearest cent, the nearest whole number or some other accuracy.

Part 3 of 6

Doubling, halving and other factors

Some questions give the multiplier in words, not as a percentage. 'Doubles every 20 minutes' means $k = 2$ with a period of 20 minutes. 'Halves every 6 hours' means $k = 0.5$ with a period of 6 hours. 'Triples' means $k = 3$ .
The method does not change: count the periods, then use that number as the power. 300 bacteria that double every 20 minutes become $300 \times 2^{6} = 19 200$ after 2 hours, because 2 hours is 6 periods of 20 minutes.
The trap is to use the time as the power. $300 \times 2^{2}$ treats 2 hours as 2 doublings, but in 2 hours the number doubles 6 times.
If the time is $t$ minutes, the number of 20-minute periods is $\frac{t}{20}$ . So a formula for the number of bacteria after $t$ minutes is:
$N = 300 \times 2^{t / 20}$
Halving is decay with $k = 0.5$ . 80 mg of a substance that halves every 6 hours leaves $80 \times {0.5}^{4} = 5$ mg after 24 hours, because 24 hours is 4 periods of 6 hours.
In this model the amount is never zero. Halving 5 gives 2.5, then 1.25, then 0.625. Half of a positive number is still positive, so the value gets closer and closer to zero but never reaches it.

Part 4 of 6

Working backwards: the start value and the rate

Sometimes you know the value after $n$ periods and you need the start value. The forward step was $V = V_{0} \times k^{n}$ . To undo a multiplication, divide:
$V_{0} = \frac{V}{k^{n}}$
Example: after 3 years of 5% growth per year, an account holds $9261. The start amount was $9261 \div {1.05}^{3} = 8000$ dollars. Check forwards: $8000 \times {1.05}^{3} = 9261$ .
Divide by all of $k^{n}$ . The value was multiplied by $1.05$ three times, so you undo all three. Dividing by $1.05$ once, or by $1.05 \times 3$ , does not undo it.
Sometimes you know the start value and the value after $n$ periods, and you need the rate. Divide to find $k^{n}$ , then take the $n$ th root to find $k$ :
$k^{n} = \frac{V}{V_{0}} \Rightarrow k = \sqrt[n]{\frac{V}{V_{0}}}$
Example: an antique worth $2500 is worth $3025 two years later. Then $k^{2} = \frac{3025}{2500} = 1.21$ , so $k = \sqrt{1.21} = 1.1$ .
Then turn $k$ back into a percentage. For growth the rate is $(k - 1) \times 100$ : here $0.1 \times 100 = 10 %$ per year. For decay the rate is $(1 - k) \times 100$ .
A decay example: a car's value falls from $24 000 to $15 000 in 4 years. So $k^{4} = \frac{15 000}{24 000} = 0.625$ and $k = \sqrt[4]{0.625} = 0.88913 \dots$ The value falls by $(1 - 0.88913 \dots) \times 100 = 11.1 %$ per year, to 3 significant figures.
The trap is to share the total change equally between the years. The antique gained 21% in 2 years, but that is not 10.5% per year. Check: ${1.105}^{2} = 1.221025$ , a gain of more than 22%, which is too much. Each year's growth is on top of the growth before, so the yearly rate is less than the total divided by the number of years.

Part 5 of 6

How many periods until a target?

A question can ask: 'After how many complete years is the car first worth less than $9000?' The answer is a whole number of periods. You find it by trying whole numbers.
Work out the value for whole numbers of years, and write each one down. For the car that costs $18 500 and loses 12% each year:
$18 500 \times {0.88}^{5} = 9763.04 \dots 18 500 \times {0.88}^{6} = 8591.47 \dots$
After 5 years the car is still worth more than $9000. After 6 years it is worth less. So the answer is 6 years.
Write down both of those values: the last one on the wrong side of the target and the first one on the right side. Together they show that 6 is the smallest whole number that works. One value on its own does not show that.
You can also use a graph. On a graphic display calculator, draw $y = 18 500 \times {0.88}^{x}$ and $y = 9000$ , and find where they meet. The graphs meet at $x = 5.64$ , to 3 significant figures.
That $x$ is not a whole number. The value only changes at the end of each year, so the answer is the next whole number above $5.64$ , which is 6. Round up, then check with the two values. Rounding to the nearest whole number can give the wrong year.
Subtopic 3.7 gives a third method, with logarithms. The equation ${0.88}^{x} = \frac{9000}{18 500}$ has the solution $x = \frac{\log (9000 \div 18 500)}{\log 0.88} = 5.64$ , the same value as the graph.

Part 6 of 6

Graphs and tables of exponential change

The graph of $y = a \times k^{x}$ crosses the $y$ -axis at $(0, a)$ , because $k^{0} = 1$ . So the $y$ -intercept is the start value.
When $k > 1$ the curve rises and gets steeper and steeper. Each step of 1 in $x$ multiplies $y$ by $k$ , so each rise is bigger than the one before it.
When $0 < k < 1$ the curve falls and gets flatter. It gets closer and closer to the $x$ -axis but never touches it, because the value is never zero. The line $y = 0$ is an asymptote of the curve.
graph diagram — not drawn yet
One pair of axes with x from 0 to 5 and y from 0 to 800. A solid curve y = 100 × 1.5^x through (0, 100), (1, 150), (2, 225), (3, 337.5), (4, 506.25) and (5, 759.375), labelled 'growth: k = 1.5'. A second solid curve y = 400 × 0.6^x through (0, 400), (1, 240), (2, 144), (3, 86.4), (4, 51.84) and (5, 31.104), labelled 'decay: k = 0.6'. Dots mark the y-intercepts (0, 100) and (0, 400). The x-axis is labelled 'y = 0, the asymptote of the decay curve'.
Each curve meets the y-axis at its start value. The growth curve gets steeper; the decay curve gets flatter and comes close to the x-axis without touching it.
To test a table with equal steps in $x$ , divide each value by the one before it. If the ratio is the same every time, the change is exponential, and that ratio is $k$ .
Example: $y$ = 5, 15, 45, 135 for $x$ = 0, 1, 2, 3. The ratios are $15 \div 5 = 3$ , $45 \div 15 = 3$ and $135 \div 45 = 3$ . The ratio is constant, so the change is exponential with $a = 5$ and $k = 3$ : $y = 5 \times 3^{x}$ .
Compare $y$ = 5, 15, 25, 35. The differences are all 10, but the ratios are $3$ , then $\frac{5}{3}$ , then $1.4$ . A constant difference with a ratio that changes is linear change, not exponential.
Check every ratio, not only the first one. A table where the first two ratios match and the third does not is not exponential.

Worked examples

Example 1. A van costs $32 000 when it is new. Its value depreciates by 18% per year. (a) Calculate the value of the van after 4 years. Give your answer correct to the nearest dollar. (b) Find the number of complete years after which the value of the van is first less than $10 000.

Show the working

(a) An 18% decrease leaves 82% of the value each year, so the multiplier is $0.82$ . Four years is 4 periods, so the power is 4.
$32 000 \times {0.82}^{4} = 14 467.896 \dots$
To the nearest dollar the value is $14 468.
Compare the wrong, linear method: 18% of $32 000 is $5760, and $32 000 - 4 \times 5760 = 8960$ . That takes 18% of the new price every year. The real fall each year is 18% of a smaller value, so the van keeps more of its value.
(b) Try whole numbers of years and write each value down.
$32 000 \times {0.82}^{5} = 11 863.67 \dots 32 000 \times {0.82}^{6} = 9728.21 \dots$
After 5 years the van is still worth more than $10 000. After 6 years it is worth less. Both values on the page show that 6 is the smallest number of years.
Check with a graph: $y = 32 000 \times {0.82}^{x}$ meets $y = 10 000$ at $x = 5.86$ . Round up to the next whole year: 6.

Answer: (a) $14 468 (b) 6 years

Example 2. The population of a town was 42 000 in 2020. A model says the population grows by 2.5% each year. (a) Use the model to estimate the population in 2026. (b) Use the model to estimate the population in 2017. Give each answer correct to the nearest hundred.

Show the working

(a) The multiplier for a 2.5% increase is $1.025$ . From 2020 to 2026 is 6 years, so the power is 6.
$42 000 \times {1.025}^{6} = 48 707.1 \dots$
To the nearest hundred, the estimate is 48 700.
(b) 2017 is 3 years before 2020. The 2020 population is the 2017 population multiplied by ${1.025}^{3}$ , so divide to go back.
$42 000 \div {1.025}^{3} = 39 001.1 \dots$
To the nearest hundred, the estimate is 39 000. Check forwards: $39 001.1 \dots \times {1.025}^{3} = 42 000$ .
Both answers are estimates. The model assumes the rate stays at 2.5% every year, and a real population does not have to do that.

Answer: (a) 48 700 (b) 39 000

Example 3. (a) A culture of 450 cells triples every 4 hours. Work out the number of cells after 1 day. (b) Write a formula for the number of cells, $N$ , after $t$ hours. (c) 96 g of a substance halves in mass every 5 days. Work out the mass left after 30 days.

Show the working

(a) 'Triples' means $k = 3$ . One day is 24 hours, and $24 \div 4 = 6$ , so there are 6 periods.
$450 \times 3^{6} = 450 \times 729 = 328 050$
Using 24 as the power would be the trap: the number triples 6 times in a day, not 24 times.
(b) After $t$ hours the number of 4-hour periods is $\frac{t}{4}$ . That is the power.
$N = 450 \times 3^{t / 4}$
Check with (a): when $t = 24$ , the power is $\frac{24}{4} = 6$ , which gives $328 050$ again.
(c) 'Halves' means $k = 0.5$ . 30 days is $30 \div 5 = 6$ periods.
$96 \times {0.5}^{6} = \frac{96}{64} = 1.5$
The mass is 1.5 g. The units stay the same as the start value: grams in, grams out.

Answer: (a) 328 050 cells (b) $N = 450 \times 3^{t / 4}$ (c) 1.5 g

Example 4. A painting was worth $40 000 in 2019 and $46 225 in 2021. Its value grows by the same percentage each year. (a) Find this percentage. (b) Estimate the value of the painting in 2025. (c) Estimate its value in 2018. Give the answers to (b) and (c) correct to the nearest dollar.

Show the working

(a) From 2019 to 2021 is 2 years, so the value was multiplied by $k^{2}$ .
$k^{2} = \frac{46 225}{40 000} = 1.155625$
Take the square root: $k = \sqrt{1.155625} = 1.075$ . The rate is $(1.075 - 1) \times 100 = 7.5 %$ per year.
Do not share the total out. The total gain is 15.5625% in 2 years, but the yearly rate is 7.5%, not about 7.8%, because the second year's growth is on a bigger value.
(b) Start from the 2019 value. From 2019 to 2025 is 6 years.
$40 000 \times {1.075}^{6} = 61 732.06 \dots$
To the nearest dollar, the estimate is $61 732. Use the exact multiplier $1.075$ , not a rounded one.
(c) 2018 is 1 year before 2019, so divide by the multiplier once.
$40 000 \div 1.075 = 37 209.30 \dots$
To the nearest dollar, the estimate is $37 209.

Answer: (a) 7.5% per year (b) $61 732 (c) $37 209

Where people go wrong

Mistake: Taking the same amount off every year: for a $18 500 car that depreciates by 12% per year, working out $18 500 - 3 \times 2220 = 11 840$ for its value after 3 years.

Instead: Multiply by the multiplier once for each year: $18 500 \times {0.88}^{3} = 12 607.23$ .

Each year's 12% is 12% of the value at the start of that year, which is smaller every year. Taking 12% of the new price every year is linear change, a different model.

Mistake: Using the percentage itself as the multiplier, e.g. $18 500 \times {0.12}^{3}$ for a 12% decrease, or using $1.12$ for a decrease.

Instead: A decrease of $r %$ keeps $(100 - r) %$ , so the multiplier is $1 - \frac{r}{100}$ : here $0.88$ . An increase uses $1 + \frac{r}{100}$ .

$0.12$ is the part that is lost, not the part that is kept. $18 500 \times {0.12}^{3}$ is about $32, which is clearly not the value of a 3-year-old car.

Mistake: Using the time as the power instead of the number of periods, e.g. $300 \times 2^{2}$ for bacteria that double every 20 minutes for 2 hours.

Instead: Divide the time by the length of one period first: $120 \div 20 = 6$ , so $300 \times 2^{6} = 19 200$ .

The power counts how many times you multiply by $k$ . That is the number of periods, not the number of hours.

Mistake: Finding a yearly rate by sharing the total change between the years, e.g. a 21% increase in 2 years called 10.5% per year.

Instead: Find $k^{n}$ from the two values, then take the $n$ th root: $k^{2} = 1.21$ , $k = 1.1$ , so the rate is 10% per year.

Each year's growth is on top of the growth before it. Shared-out rates give too much growth when you check them forwards.

Mistake: Rounding a graph or logarithm answer to the nearest whole number for 'how many complete years', e.g. giving 8 years when $5000 at 6% per year first passes $8000 at $x = 8.07$ .

Instead: Round up to the next whole number, then check with values: $5000 \times {1.06}^{8} = 7969.24 \dots$ is still below $8000 and $5000 \times {1.06}^{9} = 8447.39 \dots$ is above it, so the answer is 9 years.

The target is not passed until the curve crosses it. Any part of a year after the last whole period still needs one more whole period.

Check yourself

Write down the multiplier for (a) growth of 6% per year, (b) decay of 6% per year, (c) decay of 0.5% per month.
Answer
(a) $1.06$ (b) $0.94$ (c) $0.995$
Without a calculator, work out the value of $1000 after 2 years if it grows by 10% per year.
Answer
$1210. $1000 \times 1.1 = 1100$ , then $1100 \times 1.1 = 1210$ . Linear growth of $100 per year would give only $1200.
A phone costs $800. It loses 30% of its value each year. Find its value after 2 years.
Answer
$392. The multiplier is $0.7$ , and $800 \times {0.7}^{2} = 800 \times 0.49 = 392$ .
A population of 2400 birds decreases by 8% each year. Find the number of birds after 5 years, correct to the nearest whole number.
Answer
1582. $2400 \times {0.92}^{5} = 1581.79 \dots$ , which rounds to 1582.
640 mg of a medicine is in a patient's blood. The amount halves every 4 hours. Find the amount left after 12 hours.
Answer
80 mg. 12 hours is $12 \div 4 = 3$ periods, and $640 \times {0.5}^{3} = 80$ .
For $x$ = 0, 1, 2, 3 the values of $y$ are 5, 15, 45, 135. Determine whether $y$ changes exponentially. If it does, find a formula for $y$ .
Answer
Yes. Every ratio is 3 ( $15 \div 5$ , $45 \div 15$ , $135 \div 45$ ), and $y = 5$ when $x = 0$ , so $y = 5 \times 3^{x}$ .
A machine was worth $50 000. Two years later it was worth $40 500. Its value fell by the same percentage each year. Find that percentage.
Answer
10% per year. $k^{2} = \frac{40 500}{50 000} = 0.81$ , so $k = \sqrt{0.81} = 0.9$ , and $(1 - 0.9) \times 100 = 10$ .
After 3 years of 5% compound growth per year, an account holds $9261. Find the amount at the start.
Answer
$8000, because $9261 \div {1.05}^{3} = 8000$ . Check: $8000 \times {1.05}^{3} = 9261$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Ravi buys a motorbike for $6400. The value of the motorbike decreases by 15% each year.

(a) Calculate the value of the motorbike after 4 years. Give your answer correct to the nearest dollar. 2 marks

Mark scheme

M1 6400 × 0.85⁴ oe
A1 3341 cao

(b) Find the number of complete years after which the value of the motorbike is first less than $2000. 3 marks

Mark scheme

M1 6400 × 0.85ⁿ evaluated for at least one whole number n ≥ 6, or 0.85ⁿ < 0.3125 oe, or graphs of y = 6400 × 0.85ˣ and y = 2000 drawn
M1 2051.69… (awrt 2052) for 7 years and 1743.93… (awrt 1744) for 8 years, both seen; or 7.157… (awrt 7.16) from a graph or logarithms
A1 8 cao

(c) Sara's motorbike loses value at the same rate. After 2 years it is worth $3757. Calculate the price Sara paid for her motorbike. 2 marks

Mark scheme

M1 3757 ÷ 0.85² oe, or 0.7225x = 3757
A1 5200 cao

Question 2. The population, $P$ , of a town $t$ years after 2020 is modelled by $P = 24 000 \times {1.03}^{t}$ .

(a) Write down the population of the town in 2020. 1 mark

Mark scheme

B1 24 000

(b) Write down the percentage increase in the population each year. 1 mark

Mark scheme

B1 3 (%)

(c) Calculate the population in 2030. Give your answer correct to the nearest hundred. 2 marks

Mark scheme

M1 24 000 × 1.03¹⁰ oe
A1 32 300 cao

(d) The population, $Q$ , of a second town is modelled by $Q = 40 000 \times {0.98}^{t}$ . Find the smallest whole number value of $t$ for which $P$ is greater than $Q$ . 3 marks

Mark scheme

M1 P and Q both evaluated for the same whole number t ≥ 9, or 24 000 × 1.03ᵗ = 40 000 × 0.98ᵗ oe, or graphs of P and Q drawn
M1 for t = 10: 32 254 and 32 683 (awrt) and for t = 11: 33 222 and 32 029 (awrt), all seen; or 10.27 (awrt 10.3) from graphs or logarithms
A1 11 cao

Question 3.

(a) A colony of 500 cells doubles in size every 3 hours. Work out the number of cells after 12 hours. 2 marks No calculator

Mark scheme

M1 500 × 2⁴ oe, or 4 doublings shown, e.g. 1000, 2000, 4000
A1 8000 cao

(b) The value of a savings account at the start of each year is $2000, $2200, $2400 and $2600. Explain whether the value is growing exponentially. 2 marks No calculator

Mark scheme

M1 two ratios compared, e.g. 2200 ÷ 2000 = 1.1 and 2400 ÷ 2200 = 1.09…, or 2200 × 1.1 = 2420 ≠ 2400, or the equal differences of 200 identified
A1 not exponential, with a reason: the ratio of consecutive values is not constant, or it increases by the same amount (200) each year oe; dep on M1

(c) Sketch the graph of $y = 4 \times {0.5}^{x}$ for $x \geq 0$ . 2 marks No calculator

Mark scheme

B1 decreasing curve that gets less steep and stays above the x-axis without touching it
B1 y-intercept at 4 marked or labelled

How this is examined

The number of whole periods before a value first passes a target: A Find part in a depreciation or growth context, worth about 3 marks, after a part that asks for the value at a given time Values written down for the last whole period on the wrong side of the target and the first period on the right side, e.g. $6400 \times {0.85}^{7} = 2051.69$ and $6400 \times {0.85}^{8} = 1743.94$ for a target of $2000, then the whole number, 8. A graph or logarithm value such as $x = 7.16$ is not the answer until it is rounded up to the next whole number.
Working backwards to the start value, or to the rate per period: A 2- or 3-mark Calculate or Find in which the value after $n$ periods is given For the start value: the division by the multiplier to the power $n$ , e.g. $3757 \div {0.85}^{2}$ , then $5200$ . For the rate: $k^{n} = \frac{V}{V_{0}}$ written down, then the $n$ th root, e.g. $\sqrt{1.21} = 1.1$ , then $k$ turned into a percentage, 10% per year. Total change divided by the number of years is not a valid method.
The value after $n$ periods as start value times the multiplier to the power $n$ : A 2-mark Calculate in a depreciation, population or interest context The calculation with its numbers, e.g. $6400 \times {0.85}^{4}$ , with $0.85$ (not $0.15$ ) for a 15% decrease and the number of periods as the power. Then the value to the accuracy the question asks for.
Deciding from a table whether a change is exponential or linear: A 2-mark Explain, from a table of values at equal steps in time The ratios of consecutive values worked out and compared (or the equal differences identified), then the conclusion with its reason: a constant ratio means exponential, a constant difference means linear.
The shape and start point of an exponential growth or decay graph: A 2-mark Sketch of a graph such as $y = 4 \times {0.5}^{x}$ The y-intercept marked at the start value, and the correct shape: decay falls and flattens towards the x-axis without touching or crossing it; growth rises more and more steeply.
Reading the start value and the rate from a model such as $P = 24 000 \times {1.03}^{t}$ : 1-mark Write down parts at the start of a longer modelling question The number in front of the power is the value when $t = 0$ . The rate is $(k - 1) \times 100$ for growth and $(1 - k) \times 100$ for decay: $1.03$ gives 3% growth, $0.98$ gives a 2% decrease.

Words to know

Exponential growth: An increase by the same multiplier $k > 1$ in each equal period, so the increases get bigger over time.
Exponential decay: A decrease by the same multiplier $0 < k < 1$ in each equal period, so the decreases get smaller and the value never reaches zero.
Linear change: A change by the same amount in each equal period. It is not exponential.
Multiplier: The number $k$ you multiply by in each period: $1.025$ for growth of 2.5%, $0.88$ for decay of 12%, $2$ for doubling, $0.5$ for halving.
Period: The length of time between one multiplication and the next, such as a year, a month or 20 minutes.
Start value: The value $V_{0}$ at time zero. In $V = V_{0} \times k^{n}$ it is the number in front of the power, and on the graph it is the $y$ -intercept.
Depreciation: A fall in the value of an item, such as a car or a machine, over time.
Appreciation: A rise in the value of an item, such as a painting, over time.
Asymptote: A line that a curve gets closer and closer to without touching. For $y = a \times k^{x}$ it is the line $y = 0$ .

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