Writing an exponential function from a table or graph
A table of values gets multiplied by the same number at every step. How do you turn the table back into the rule that made it?
Where you are 6 in Exponential functions
When you reach for it
You have some outputs of an exponential function, in a table or as points on a graph, and you want the rule
that produced them. With the rule you can find any output, not only the listed ones, and its two numbers say what is going on: a is the starting value and b is the growth or decay factor, the number the output is multiplied by for each step of 1 in x.
Reach for it also before you know the data are exponential. The arithmetic that finds b, dividing each output by the one before, is the same arithmetic that tells you whether an exponential rule fits at all.
The procedure
Everything rests on two facts about y = a \cdot b^x, where a \neq 0, b > 0 and b \neq 1.
At x = 0 the output is a. Any nonzero number to the power 0 is 1, so y = a \cdot b^0 = a \cdot 1 = a. On a graph, that is the height where the curve crosses the y-axis.
A step of 1 in x multiplies the output by b. Raising the exponent by 1 adds one more factor of b:
So dividing any output by the one before it gives b, and gives the same b every time.
A table with steps of 1, starting at 0
Read a from the row x = 0. Divide each output by the one before. If every quotient is the same number, that number is b. Checking all the quotients is part of the work, not a formality: any two numbers have a quotient, so one quotient proves nothing, and only agreement across the whole table shows the factor is constant.
A table that does not start at 0
Find b the same way, then walk back to x = 0. A step forward multiplies by b, so a step back divides by b. Dividing by a decay factor such as \displaystyle \frac{1}{4} makes the number larger, which is right: something that shrinks as x grows was bigger earlier.
A table with steps of 2
If x goes up by 2 from row to row, each row multiplies by b twice:
so the quotient of neighboring outputs is b^2, not b, and b is the positive square root of it. The negative root is never a candidate, because the base must be positive: a negative base would flip the sign of the output at every step, and at \displaystyle x = \frac{1}{2} it would give no real number at all.
Two points
Two points on the curve give two equations, one for each. Divide the larger-x equation by the other. The a cancels, and the quotient of the outputs is b raised to the gap between the x-values. That is the steps-of-2 reasoning again, written with letters. Once b is known, either equation gives a.
Reading a graph
Use points whose coordinates are labeled, or that sit exactly where grid lines cross. A height read by eye off a curve is an estimate, and an estimate of b is then raised to a power, which magnifies the error. The best pair is (0, a) and (1, a \cdot b): the first gives a outright, and the second divided by the first gives b. The shape of the curve tells you in advance what to expect: a rising curve has b > 1, a falling one has 0 < b < 1.
Not a sequence formula
A geometric sequence is written a_n = a_1 \cdot r^{n-1}, because its counting starts at n = 1 and the first term must come with the exponent 0. A function’s x starts at 0, and a is the output at x = 0, so the exponent is plain x. When a table starts at x = 1, its first output is a \cdot b, not a.
Exponential, linear or neither
With equal steps in x, a linear function adds the same amount at every step and an exponential function multiplies by the same factor. So compute both the differences and the quotients of neighboring outputs. Constant differences mean linear; constant quotients mean exponential; neither constant means neither family fits. The wider question of which family suits a situation, quadratics included, is Choosing a function family for data.
Worked examples
A coral patch, steps of 1 from 0
A diver measures the area of one coral patch on a reef each year. The numbers are made up for this example.
| t (years) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| A (m²) | 20 | 24 | 28.8 | 34.56 |
Write a rule for A, and use it to predict the area in year 5.
SolutionThe row t = 0 gives the starting value, a = 20. Now divide each area by the one before:
All three agree, so the area is multiplied by b = 1.2 each year, a growth of 20\% a year, and
Check one row: 20(1.2)^3 = 20 \times 1.728 = 34.56. For year 5, A = 20(1.2)^5 = 20 \times 2.48832 = 49.7664, about 49.8 square meters, if the growth keeps up.
A table that starts at 1
Write a rule for this table.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| y | 48 | 12 | 3 | 0.75 |
The quotients are \displaystyle \frac{12}{48} = \frac{1}{4}, \displaystyle \frac{3}{12} = \frac{1}{4} and \displaystyle \frac{0.75}{3} = \frac{1}{4}, so \displaystyle b = \frac{1}{4}.
The table has no row x = 0, and 48 is not a: it is the output after one step, a \cdot b. Step back once by dividing by b:
So \displaystyle y = 192\left(\frac{1}{4}\right)^x. Check the last row: \displaystyle 192\left(\frac{1}{4}\right)^4 = \frac{192}{256} = 0.75.
Tinted film, steps of 2
Wynford shines a lamp at a light meter and stacks sheets of the same tinted film in front of it. With n sheets, the meter reads L.
| n (sheets) | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| L | 500 | 320 | 204.8 | 131.072 |
What fraction of the light does one sheet let through? Write a rule for L, and find the reading with one sheet.
SolutionThe quotients are \displaystyle \frac{320}{500} = 0.64, \displaystyle \frac{204.8}{320} = 0.64 and \displaystyle \frac{131.072}{204.8} = 0.64. They agree, but each covers two sheets, so
taking the positive root because a base must be positive. Each sheet lets through 80\% of the light that reaches it. The row n = 0 gives a = 500, so
With one sheet, L = 500 \times 0.8 = 400. Writing L = 500(0.64)^n would be wrong: at n = 2 it gives 500 \times 0.4096 = 204.8, the reading for four sheets.
From a graph through (0, 8)
Write the rule for the curve in the figure.
The curve crosses the y-axis at the labeled point (0, 8), so a = 8. One step on, the labeled point is (1, 6), which is a \cdot b:
So y = 8(0.75)^x. The third labeled point checks it: 8(0.75)^2 = 8 \times 0.5625 = 4.5. The curve falls, and b is between 0 and 1, as it should be.
Two points, neither at 0
(a) Write the rule for the curve through the labeled points (1, 6) and (3, 54).
(b) An exponential function y = a \cdot b^x passes through (2, 81) and (4, 65.61). Find its rule.
Solution(a) The points give a \cdot b = 6 and a \cdot b^3 = 54. Divide the second by the first, and a cancels:
So b = 3, the positive root. Then a \cdot 3 = 6 gives a = 2, and y = 2 \cdot 3^x. Check: 2 \cdot 3^3 = 54.
(b) The gap in x is again 2, so
From a \cdot (0.9)^2 = 81, \displaystyle a = \frac{81}{0.81} = 100. So y = 100(0.9)^x. Check: 100(0.9)^4 = 100 \times 0.6561 = 65.61.
Exponential, linear or neither
Decide whether each table fits a linear rule, an exponential rule or neither, and write the rule when there is one.
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| table P | 4 | 11 | 18 | 25 |
| table Q | 125 | 75 | 45 | 27 |
| table R | 1 | 3 | 9 | 19 |
For each table, compute the differences and the quotients of neighboring outputs.
Table P: the differences are 7, 7, 7. The quotients, \displaystyle \frac{11}{4} = 2.75 and \displaystyle \frac{18}{11} \approx 1.64, already disagree. Adding 7 each time is linear: y = 7x + 4.
Table Q: the differences are -50, -30, -18, not constant. The quotients are \displaystyle \frac{75}{125} = \frac{3}{5}, \displaystyle \frac{45}{75} = \frac{3}{5} and \displaystyle \frac{27}{45} = \frac{3}{5}, all equal. It is exponential with a = 125: \displaystyle y = 125\left(\frac{3}{5}\right)^x.
Table R: the differences are 2, 6, 10, not constant. The quotients are 3, 3 and \displaystyle \frac{19}{9} \approx 2.11: the first two agree and the third does not, which is why every quotient has to be checked. Neither a linear nor an exponential rule fits. (Its differences grow by 4 each time, the mark of a quadratic; in fact y = 2x^2 + 1.)
Boundaries
- The form is y = a \cdot b^x with a \neq 0, b > 0, b \neq 1. Its outputs are then never 0 and all have the same sign as a. A table with a 0 output, or with outputs that change sign, is not of this form.
- Neighboring quotients need equal steps in x. If x runs 0, 1, 3, the quotient from x = 1 to x = 3 is b^2 while the one before is b, and the two will not match even for an exponential. Over a gap of 3 the quotient is b^3, and b is its cube root: b^3 = 8 gives b = 2.
- The curve’s asymptote must be the x-axis. A shifted curve such as y = 3 \cdot 2^x + 1 levels off at y = 1, not y = 0 (see What an asymptote is). It crosses the y-axis at 4, not at 3, and its quotients are not constant. Rules of that kind are Transforming an exponential graph.
- Measured data are rarely exact. Quotients that are close but not equal mean an exponential rule is at best a model. Choosing the best-fitting rule for such data is a later topic.
- The rule speaks for the range it came from. A coral patch cannot grow by 20\% a year for ever; predicting far beyond the table assumes the factor holds, which only the situation can say.
Where this goes next
Linear versus exponential growth puts a linear and an exponential rule side by side, and shows that exponential growth always overtakes linear growth in the end. Building an exponential model from two points does the two-point work for any gap and any data, solving equations rather than reading quotients. And when a question asks for x (in which year does the coral reach 100 square meters?), the unknown sits in the exponent; Logarithms are the tool that brings it down.
Exercises
Tables that start at 0
Write a rule y = a \cdot b^x for each.
- x = 0, 1, 2, 3 gives y = 3, 6, 12, 24.
- x = 0, 1, 2, 3 gives y = 400, 100, 25, 6.25.
- A second coral patch, damaged by a storm, shrinks each year (made-up data): t = 0, 1, 2, 3 gives A = 80, 76, 72.2, 68.59 square meters. Write the rule, and say by what percent it shrinks each year.
Tables that start elsewhere
- x = 1, 2, 3, 4 gives y = 10, 20, 40, 80.
- x = 2, 3, 4, 5 gives y = 50, 250, 1250, 6250.
- x = 3, 4, 5, 6 gives y = 8, 4, 2, 1.
Steps of 2
- x = 0, 2, 4, 6 gives y = 7, 28, 112, 448.
- x = 0, 2, 4 gives y = 250, 90, 32.4.
- Through a darker film, the meter reads 300 with one sheet and 75 with three sheets. What fraction of the light does each sheet let through, and what does the meter read with no film? Write the rule.
From a graph or two points
- Write the rule for the curve in the figure.
- An exponential curve y = a \cdot b^x passes through (1, 12) and (3, 108). Write its rule.
- The same, for a curve through (2, 100) and (4, 25).
- The same, for a curve through (-1, 2) and (1, 18).
Exponential, linear or neither
Decide which, and write the rule when there is one.
- x = 0, 1, 2, 3 gives y = 5, 15, 45, 135.
- x = 0, 1, 2, 3 gives y = 7, 4, 1, -2.
- x = 0, 1, 2, 3 gives y = 2, 3, 5, 8.
Reasoning
- Explain why, when x goes up in steps of 2, the quotient of neighboring outputs is b^2, and why you then take only the positive square root.
- Ximena is given x = 1, 2, 3, 4 with y = 8, 32, 128, 512, and writes y = 8 \cdot 4^x because the first output is 8. Explain what went wrong, and write the correct rule.
Answers
- y = 3 \cdot 2^x. The row x = 0 gives a = 3, and every quotient is 2.
- y = 400(0.25)^x. a = 400, and every quotient is \displaystyle \frac{100}{400} = 0.25.
- A = 80(0.95)^t. Every quotient is 0.95, so it shrinks by 5\% a year.
- y = 5 \cdot 2^x. b = 2, and a = 10 \div 2 = 5.
- y = 2 \cdot 5^x. b = 5; stepping back twice, a = 50 \div 5 \div 5 = 2.
- \displaystyle y = 64\left(\frac{1}{2}\right)^x. \displaystyle b = \frac{1}{2}; stepping back three times multiplies by 2 three times, a = 8 \times 8 = 64.
- y = 7 \cdot 2^x. \displaystyle b^2 = \frac{28}{7} = 4, so b = 2, and a = 7.
- y = 250(0.6)^x. \displaystyle b^2 = \frac{90}{250} = 0.36, so b = 0.6, and a = 250.
- Each sheet lets through half, b = 0.5; with no film the meter reads 600; L = 600(0.5)^n. Two sheets separate the readings, so \displaystyle b^2 = \frac{75}{300} = 0.25 and b = 0.5. Stepping back one sheet, a = 300 \div 0.5 = 600.
- y = 2(2.5)^x. The curve crosses at (0, 2), so a = 2, and \displaystyle b = \frac{5}{2} = 2.5.
- y = 4 \cdot 3^x. \displaystyle b^2 = \frac{108}{12} = 9, so b = 3, and a = 12 \div 3 = 4.
- \displaystyle y = 400\left(\frac{1}{2}\right)^x. \displaystyle b^2 = \frac{25}{100} = \frac{1}{4}, so \displaystyle b = \frac{1}{2}, and \displaystyle a = 100 \div \frac{1}{4} = 400.
- y = 6 \cdot 3^x. The gap is 2, so \displaystyle b^2 = \frac{18}{2} = 9 and b = 3; then a = 18 \div 3 = 6. Check: 6 \cdot 3^{-1} = 2.
- Exponential, y = 5 \cdot 3^x. Every quotient is 3; the differences 10, 30, 90 are not equal.
- Linear, y = -3x + 7. Every difference is -3; the quotients do not even keep a sign.
- Neither. The differences are 1, 2, 3 and the quotients are 1.5, \displaystyle \frac{5}{3} \approx 1.67 and 1.6.
- Each step of 2 in x adds two factors of b to the exponent, a \cdot b^{x+2} = (a \cdot b^x) \cdot b^2, so neighboring outputs differ by the factor b^2. Both \displaystyle \sqrt{b^2} and \displaystyle -\sqrt{b^2} square to b^2, but the base of an exponential function must be positive, so only the positive root is possible.
- y = 2 \cdot 4^x. The first output is at x = 1, so 8 is a \cdot b, not a; the factor is 4, and stepping back gives a = 8 \div 4 = 2. Ximena’s rule gives 32 at x = 1, four times too much at every row.