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Build linear and exponential functions from situations, and compare how the two kinds of growth behave over the long run.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you learn to turn a described situation into a function, deciding first whether each step adds a fixed amount — a linear model — or multiplies by a fixed factor, which is exponential. You then compare the two: why a percentage change is a multiplier rather than an addition, and why an exponential function eventually passes any linear one, however steep.
You already write $y = mx + b$ from a starting value and a rate, and you have just met geometric sequences, where each step multiplies. Building a model is those two skills plus one question asked before either of them: does this situation add the same amount each step, or multiply by the same factor?
Model: a function written to describe a real situation.
Initial value: the amount at time zero — the $b$ of $y = mx + b$, and the $a$ of $y = ab^t$.
Rate of change: how much is added per step, in a linear model.
Growth factor: what the quantity is multiplied by each step. A rise of $5\%$ gives $1.05$; a fall of $20\%$ gives $0.8$.
Two models cover an enormous amount of the world, and one question separates them. If the situation changes by the same amount each step, it is linear: $y = b + mt$, where $b$ is the value at the start and $m$ is the amount added (or, if negative, taken away) per step. A tank of $500$ litres draining at $20$ litres a minute is $V = 500 - 20t$. If instead the situation changes by the same percentage each step, it is exponential: $y = a \cdot b^t$, where $a$ is the starting value and $b$ is the growth factor. A rise of $5\%$ a year means keeping all of it and gaining a twentieth, so $b = 1.05$ and $P = 800(1.05)^t$. Anything that doubles, halves or is repeatedly discounted is exponential, and the exponent counts periods, not units of time.
Another way: diagram
Two tables side by side for $t = 0, 1, 2, 3$: the left one $500, 480, 460, 440$ with $-20$ arcs between the rows; the right one $800, 840, 882, 926.1$ with $\times 1.05$ arcs.
Another way: story
Ask what happens between two consecutive rows of a table. Subtract them: if the answer is always the same, the model is linear. Divide them: if that answer is always the same, it is exponential.
"$5\%$ growth means multiply by $0.05$." That would throw away everything the quantity already had. It keeps its $100\%$ and gains $5\%$: multiply by $1.05$.
"A quantity that falls must be linear." A car losing $20\%$ of its value a year loses a different number of dollars every year. That is exponential decay, with multiplier $0.8$.
"Doubling every $3$ hours for $12$ hours means $\times 2 \times 12$." There are $12 \div 3 = 4$ doubling periods, so the factor is $2^4 = 16$.
At $t = 0$ there are $500$ litres: that is the initial value.
Start from time zero.
Each minute removes the same $20$ litres, so the change is an addition of $-20$ per minute: linear.
Same amount each step.
So $V = 500 - 20t$.
After a year the population is $105\%$ of what it was, so the growth factor is $1.05$.
Keep the $100\%$, add the $5\%$.
Each year multiplies by $1.05$ again, so after $t$ years the factor is $1.05^t$.
Repeated multiplication.
$P = 800(1.05)^t$. After $2$ years: $800 \times 1.05 \times 1.05 = 882$.
The $\$3$ is paid once, so it is the value at $k = 0$.
Each kilometre adds the same $\$2$, so $C = 3 + 2k$.
A tank holds $532$ litres and drains at $27$ litres per minute. Write the volume $V$ left after $t$ minutes.
Answer:
A taxi charges $\$12$ to get in and $\$5$ for each kilometre. Write the cost $C$ of a journey of $k$ kilometres.
Answer:
A population grows by $24\%$ each year. What does it multiply by each year?
A colony of $80$ bacteria doubles every $5$ hours. How many are there after $25$ hours?
answer
A car worth $\$22000$ loses $12\%$ of its value every year. Which rule gives its value $V$ after $t$ years?
You have just built both kinds of model, and you can evaluate $2^x$ for small $x$. What is new is a comparison over the long run — the fact that an exponential function overtakes a linear one no matter how big the linear one's rate is, which almost nobody believes at first sight.
Growth: a multiplier greater than $1$. Decay: a multiplier between $0$ and $1$.
Eventually: from some input onwards, and for ever after. It says nothing about small inputs.
Common difference / common ratio: what stays constant between consecutive outputs of a linear / exponential relationship.
Crossing point: the input at which one function overtakes the other.
To tell a linear relationship from an exponential one in a table, look between consecutive rows: equal differences mean linear, equal ratios mean exponential. Over the long run the two behave completely differently. A linear function gains the same amount at every step; an exponential one with a factor above $1$ gains a share of itself, which grows as it does. So an exponential function eventually passes any linear function, whatever the slope — $2^x$ overtakes $100x$ somewhere near $x = 10$ and is a thousand times ahead not long afterwards. When the factor is below $1$ the same reasoning runs backwards: the quantity keeps a fixed share each step, so it falls quickly at first and then ever more slowly, approaching zero without reaching it.
Another way: diagram
A graph of $y = 10x$ as a straight line and $y = 2^x$ as a curve, with the curve below the line until they cross near $x = 6$ and shooting past it after that.
Another way: story
A linear quantity is paid a fixed wage; an exponential one earns interest on everything it already has. The wage looks better for a long time, and then it does not.
"$100x$ is bigger than $2^x$, look at the numbers." At $x = 5$, yes: $500$ against $32$. At $x = 10$: $1000$ against $1024$. At $x = 20$: $2000$ against over a million. Early values decide nothing.
"$50(0.8)^t$ is a linear decrease of $0.8$." It is a decay: the quantity is multiplied by $0.8$ each step, so it loses $20\%$ of a shrinking amount and never quite reaches zero.
"A table with growing outputs must be exponential." Test it. $3, 6, 9, 12$ grows and is linear; $3, 6, 12, 24$ grows and is exponential. Constant differences against constant ratios is the test.
At $x = 5$: $f = 50$, $g = 32$. The line is ahead.
Try small inputs first.
At $x = 6$: $f = 60$, $g = 64$. The exponential has just passed it.
The crossing point.
At $x = 10$: $f = 100$, $g = 1024$. It stays ahead and the gap widens for ever.
The multiplier $0.8$ is below $1$, so the quantity shrinks: decay.
Below one means decay.
Keeping $0.8$ of it means keeping $80\%$, so $20\%$ is lost each step.
The loss is a percentage, not a fixed amount.
Differences: $3$, then $6$, then $12$. Not constant, so not linear.
Ratios: $2$, $2$, $2$. Constant, so exponential with multiplier $2$.
Compare $f(x) = 191x$ with $g(x) = 2^x$. Which is larger once $x$ is large enough?
A quantity starts at $2$ and doubles every year. After how many years does it first exceed $32$?
Answer:
$f(x) = 20x$ and $g(x) = 2^x$. What is the smallest whole number $x \ge 1$ for which $g(x) > f(x)$?
Answer:
A table gives the outputs $4$, $16$, $64$, $256$ for the inputs $0, 1, 2, 3$. Is the relationship linear or exponential?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A quantity starts at $72$ and doubles every $8$ hours. Which kind of function models it?
What does $h(t) = 59 \times 0.7^{\,t}$ describe?
You can build a linear or exponential model from a situation and compare their growth. Explain why $5\%$ growth multiplies by $1.05$ and not by $0.05$, and why $2^x$ eventually beats $100x$.
8. Your turn: a taxi charges $\$3$ plus $\$2$ a kilometre. Write the cost $C$ for $k$ kilometres, step 2
20. Your turn: outputs $3, 6, 12, 24$ for inputs $0, 1, 2, 3$ — linear or exponential?, step 2