Graphs of exponential functions
Doubling sends numbers soaring to the right. What does doubling look like to the left, where the inputs are negative?
Where you are 6 in Exponential functions
When you reach for it
A rule such as y = 2^x or C = 5 \cdot 2^t describes a quantity multiplied by the same factor at every step, the growth and decay whose factor is studied in Growth and decay factors. A table gives a handful of values. The graph shows all of them at once: where the quantity starts, how fast it climbs or falls, and the floor it never goes through. You reach for it to read off domain and range, to say what happens far out on each side, and to compare two rules at a glance.
The procedure
Day 1: the graph of y = 2^x
A table that includes negative inputs. As for any function, start with a table. Zero and negative inputs use zero and negative exponents: 2^0 = 1 and \displaystyle 2^{-n} = \frac{1}{2^n}, so for instance \displaystyle 2^{-3} = \frac{1}{8}.
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| 2^x | \displaystyle \frac{1}{8} | \displaystyle \frac{1}{4} | \displaystyle \frac{1}{2} | 1 | 2 | 4 | 8 |
Read left to right, each output is double the one before; read right to left, each is half. The negative inputs are no special case: they are what you get by halving backward from 2^0 = 1.
Joining the points. The input x can be any real number, not only an integer. Take \displaystyle x = \frac{1}{2}: by Multiplying and dividing powers with the same base, 2^{1/2} \cdot 2^{1/2} = 2^1 = 2, so 2^{1/2} is the positive number whose square is 2, that is \displaystyle \sqrt{2} \approx 1.41, a height between 2^0 and 2^1. Every input between two integers fills in a height between its neighbors, so the plotted points are joined by one smooth, unbroken curve, not by straight segments and not left as dots.
The y-intercept. At x = 0 the output is 2^0 = 1, so the graph crosses the y-axis at (0, 1).
The horizontal asymptote. Going left, the outputs halve again and again: \displaystyle \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \ldots Half of a positive number is a smaller positive number, so the outputs never reach 0 and never go below it, yet they get as close to 0 as you like: \displaystyle 2^{-10} = \frac{1}{1024}, and 2^{-20} is under one millionth. So the x-axis, the line y = 0, is a horizontal asymptote of the graph on the left. On the right there is none: the outputs grow without bound.
Domain and range. Read as in Domain and range from a graph. The domain is all real numbers, since 2^x has a value for every x. The range is y > 0: every output is positive, and the curve climbs without a break from just above the axis to as high as you like, so it passes through every positive height, each exactly once. The height 0 is not in the range.
End behavior. In the words and arrows of End behavior, the graph rises without bound on the right and levels toward 0 on the left:
The mirror image, \displaystyle y = \left(\frac{1}{2}\right)^x. Since 1^x = 1, \displaystyle \left(\frac{1}{2}\right)^x = \frac{1}{2^x} = 2^{-x}. So the output of \displaystyle \left(\frac{1}{2}\right)^x at an input x is the output of 2^x at -x. Every point (x, y) on y = 2^x has a partner (-x, y) on \displaystyle y = \left(\frac{1}{2}\right)^x: the same height, on the other side of the y-axis. The graph of \displaystyle y = \left(\frac{1}{2}\right)^x is the graph of y = 2^x reflected in the y-axis. Its table is the table above read backward: \displaystyle 8, 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8} for x = -3 to 3.
A reflection in the y-axis keeps every height, so the y-intercept (0, 1), the asymptote y = 0, the domain and the range y > 0 are all unchanged. Left and right swap: the curve falls, and its asymptote is on the right. As x \to \infty, y \to 0; as x \to -\infty, y \to \infty.
Day 2: the family y = a \cdot b^x
An exponential function has a rule y = a \cdot b^x: a starting value a times a base b raised to the power x. In this lesson a > 0, and always b > 0 and b \ne 1.
a is the y-intercept. At x = 0, y = a \cdot b^0 = a \cdot 1 = a, so the graph crosses the y-axis at (0, a).
a stretches the graph. Each output of a \cdot b^x is a times the output of b^x at the same x. The graph of y = 3 \cdot 2^x is the graph of y = 2^x with every height tripled: (1, 2) becomes (1, 6) and \displaystyle (-1, \frac{1}{2}) becomes \displaystyle (-1, \frac{3}{2}). Three times a positive height is still positive, and three times a height near 0 is still near 0, so the asymptote stays y = 0, the range stays y > 0, and the domain stays all real numbers.
b sets the direction and the speed. Each step of 1 to the right multiplies the output by b. If b > 1 the outputs grow, so the graph rises from left to right. If 0 < b < 1 they shrink, so it falls. A larger base climbs faster on the right: for positive whole x, 3^x multiplies by 3 as many times as 2^x multiplies by 2, so y = 3^x lies above y = 2^x there. On the left the order reverses, since \displaystyle 3^{-2} = \frac{1}{9} is less than \displaystyle 2^{-2} = \frac{1}{4}: the larger base hugs the axis more closely. Every graph y = b^x passes through (0, 1), where they all meet.
Why b must be positive and not 1. With b = 1, 1^x = 1 for every x, so y = a \cdot 1^x is the horizontal line y = a: nothing grows or decays. A negative base fails differently. (-2)^x gives -2, 4, -8, 16 at x = 1, 2, 3, 4, jumping from one side of the axis to the other, and at \displaystyle x = \frac{1}{2} there is no real number whose square is -2, so (-2)^{1/2} is not a real number at all. There is no unbroken curve to draw. The base 0 fails too: 0^x = 0 for positive x, and 0^{-1} would be \displaystyle \frac{1}{0}. Looking ahead, a negative a makes every output negative and turns the graph upside down below the x-axis.
Reading a and b off a graph, a preview. If a graph passes through (0, 5) and (1, 15), then a = 5, and one step to the right multiplies by b, so \displaystyle b = \frac{15}{5} = 3. Working backward from graphs and tables in full is Writing an exponential function from a table or graph.
Graphs in a context. When the input counts time from a first measurement, the domain starts at t = 0 and the graph starts at a closed dot (0, a) instead of running off to the left. The situation may also end, and the domain with it. The range is then read from the drawn piece only.
One shift, to see the asymptote move. Adding a constant lifts every point. The graph of y = 2^x + 2 is the graph of y = 2^x raised 2 units, so its asymptote is raised too, to y = 2 (Example 7). Moving and stretching exponential graphs in general is Transforming an exponential graph.
Worked examples
Day 1: plotting y = 3^x
Make a table of y = 3^x for x = -2 to 2 and sketch the graph. Give its y-intercept, its asymptote, its domain and range, and its end behavior.
SolutionBy the meaning of a negative exponent, \displaystyle 3^{-2} = \frac{1}{3^2} = \frac{1}{9} and \displaystyle 3^{-1} = \frac{1}{3}; then 3^0 = 1, 3^1 = 3, 3^2 = 9.
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| 3^x | \displaystyle \frac{1}{9} | \displaystyle \frac{1}{3} | 1 | 3 | 9 |
Plot the points and join them with a smooth curve, since x may be any real number. The y-intercept is (0, 1), because 3^0 = 1. Each step left divides by 3, so the outputs stay positive and shrink toward 0: the asymptote is y = 0, on the left. The domain is all real numbers and the range is y > 0. As x \to \infty, y \to \infty; as x \to -\infty, y \to 0. Compared with y = 2^x it is steeper on the right (9 against 4 at x = 2) and closer to the axis on the left (\displaystyle \frac{1}{9} against \displaystyle \frac{1}{4} at x = -2).
Day 1: the mirror of y = 3^x
Use Example 1 to graph \displaystyle y = \left(\frac{1}{3}\right)^x, and describe its end behavior.
SolutionSince \displaystyle \left(\frac{1}{3}\right)^x = 3^{-x}, each point (x, y) on y = 3^x gives the point (-x, y) on the new graph: (2, 9) becomes (-2, 9), (1, 3) becomes (-1, 3), and so on.
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| \displaystyle \left(\frac{1}{3}\right)^x | 9 | 3 | 1 | \displaystyle \frac{1}{3} | \displaystyle \frac{1}{9} |
Check one point directly: \displaystyle \left(\frac{1}{3}\right)^{-2} = \frac{1}{(1/3)^2} = \frac{1}{1/9} = 9. The curve falls from left to right through (0, 1). Its asymptote is still y = 0, now on the right; the domain is all real numbers and the range y > 0. As x \to \infty, y \to 0; as x \to -\infty, y \to \infty.
Day 1: between and beyond the plotted points
(a) Estimate the height of y = 2^x at x = 1.5 from the graph, then find it exactly. (b) Does the graph reach the height 0.001? If so, between which two consecutive integers? Does it ever reach 0?
Solution(a) On the graph, x = 1.5 sits between the points (1, 2) and (2, 4), a little under 3. By Multiplying and dividing powers with the same base, \displaystyle 2^{1.5} = 2^1 \cdot 2^{1/2} = 2\sqrt{2} \approx 2.83. It is below the halfway height 3 because the curve bends upward: it climbs slowly at first and faster later.
(b) \displaystyle 2^{-9} = \frac{1}{512} \approx 0.00195, above 0.001, and \displaystyle 2^{-10} = \frac{1}{1024} \approx 0.000977, below it. The curve is unbroken and rising, so it passes the height 0.001 exactly once, between x = -10 and x = -9. Any positive height, however small, is reached somewhere further left. The height 0 is never reached, because every output is positive. Both facts together are what the range y > 0 says.
Day 2: stretching by a
Make a table of y = 3 \cdot 2^x for x = -2 to 2 and compare its graph with y = 2^x.
SolutionTriple each output of 2^x: \displaystyle 3 \cdot \frac{1}{4} = \frac{3}{4}, then \displaystyle \frac{3}{2}, 3, 6, 12.
| x | -2 | -1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| 3 \cdot 2^x | \displaystyle \frac{3}{4} | \displaystyle \frac{3}{2} | 3 | 6 | 12 |
The y-intercept is (0, 3), the value of a. The base 2 is above 1, so the graph rises. The left panel of the family figure shows both curves. The gap between them is not constant: at x = 2 it is 12 - 4 = 8, at x = -2 only \displaystyle \frac{3}{4} - \frac{1}{4} = \frac{1}{2}, because both curves close in on the same asymptote y = 0. Domain all real numbers, range y > 0, as for y = 2^x.
Day 2: duckweed on a pond
Duckweed covers 5 square meters of a pond of area 80 square meters when first measured, and its cover doubles every week (made-up numbers). Its cover after t weeks is C = 5 \cdot 2^t. Graph it, and give a sensible domain and the matching range.
SolutionDoubling from 5 gives 5, 10, 20, 40, 80 at t = 0, 1, 2, 3, 4.
At t = 4 the cover is 5 \cdot 2^4 = 80 square meters, the whole pond, so the doubling cannot go on: the rule only describes the pond up to t = 4. Time is counted from the first measurement, and the rule was never claimed for earlier times. So the domain is 0 \le t \le 4. The duckweed grows all through each week, not in jumps, so the points are joined by a curve with closed dots at both ends. On that piece C rises from 5 to 80, so the range is 5 \le C \le 80. The rule’s asymptote C = 0 belongs to negative t, outside the domain, so it plays no part here. Notice that the pond is half covered at t = 3, only one week before it is full.
Day 2: a radioactive tracer
A lab uses a radioactive tracer whose strength is 80 units at t = 0 and halves every hour (made-up numbers), so \displaystyle S = 80 \cdot \left(\frac{1}{2}\right)^t after t hours. Give the domain, range, asymptote and end behavior for t \ge 0, and find between which two whole hours the strength first falls below 1 unit.
SolutionThe base \displaystyle \frac{1}{2} is between 0 and 1, so the graph falls from the closed dot (0, 80).
The domain is t \ge 0. Every output is positive, the largest is 80 at t = 0, and the outputs get as close to 0 as you like, so the range is 0 < S \le 80. The asymptote is S = 0, and as t \to \infty, S \to 0. Halving from 80: 40, 20, 10, 5, 2.5, then 1.25 at t = 6 and 0.625 at t = 7. The strength is still above 1 at 6 hours and below it at 7, and it falls steadily, so it first drops below 1 unit between hours 6 and 7.
Day 2: watching the asymptote move
Graph y = 2^x + 2 from a table for x = -2 to 2, and give its y-intercept, asymptote and range.
SolutionAdd 2 to each output of 2^x: \displaystyle \frac{9}{4}, \frac{5}{2}, 3, 4, 6.
The y-intercept is 2^0 + 2 = 3, the point (0, 3); this rule is not of the form a \cdot b^x, so the intercept is no longer a single number in the rule. Far to the left 2^x is positive and as close to 0 as you like, so 2^x + 2 is always above 2 and as close to 2 as you like. The asymptote is the line y = 2, and the range is y > 2. As x \to -\infty, y \to 2; as x \to \infty, y \to \infty. The domain is still all real numbers.
Boundaries
- The base must satisfy b > 0 and b \ne 1. A base of 1 gives a horizontal line, and a negative base or 0 gives no unbroken graph.
- This lesson takes a > 0. With a = 0 the rule is the line y = 0; a negative a flips the graph below the axis and the range becomes y < 0.
- The asymptote y = 0 and range y > 0 belong to y = a \cdot b^x. Adding a constant moves both, as Example 7 shows.
- The smooth curve needs every real input. When the input counts whole steps only, such as generations, the graph is a set of dots. In a context the graph is drawn only over a sensible domain.
Where this goes next
Writing an exponential function from a table or graph runs this lesson backward, from labeled points to the rule. Transforming an exponential graph moves and stretches the curve in every direction, taking its asymptote along. Linear versus exponential growth sets these curves beside straight lines and shows that a rising exponential eventually overtakes any line. The idea of a line approached far out, met here as y = 0, returns in Horizontal asymptotes for many more kinds of function.
Exercises
Day 1: graphs of y = b^x
- Make a table of y = 4^x for x = -2, -1, 0, 1, 2, and give the y-intercept.
- Make a table of \displaystyle y = \left(\frac{1}{4}\right)^x for x = -2, -1, 0, 1, 2.
- Find 2^x at x = -4, x = 0 and x = 5, and write each as a point on the graph of y = 2^x.
- State the domain, range, y-intercept and horizontal asymptote of y = 5^x.
- Describe the end behavior of \displaystyle y = \left(\frac{1}{5}\right)^x in words and in arrow notation.
- The point (3, 8) is on y = 2^x. Which point on \displaystyle y = \left(\frac{1}{2}\right)^x is its mirror image? Check it in the rule.
Day 1: reading heights
- Is the point (-2, -4) on the graph of y = 2^x? Is \displaystyle \left(-2, \frac{1}{4}\right)?
- Between which two consecutive integers is the input x with 2^x = 100?
- Between which two consecutive integers is the input x with 2^x = 0.01?
Day 1: reasoning
- Jolanda says the graph of y = 2^x must cross the x-axis somewhere far to the left, since it keeps going down. Explain why it never does.
- Explain why \displaystyle y = \left(\frac{1}{2}\right)^x and y = 2^{-x} have the same graph, and use that to explain why it is the mirror image of y = 2^x.
Day 2: the rule y = a \cdot b^x
- Make a table of y = 3 \cdot 5^x for x = -2 to 2. Give the y-intercept and the asymptote.
- Make a table of \displaystyle y = 4 \cdot \left(\frac{1}{2}\right)^x for x = -2 to 2. Does the graph rise or fall? Give its end behavior.
- Say whether each graph rises or falls, and give its y-intercept: (a) y = 7(1.2)^x; (b) y = 0.5 \cdot 3^x; (c) y = 10(0.8)^x.
- Of y = 1.5^x, y = 2^x and y = 4^x, which is highest at x = 3? Which is highest at x = -3?
- Each graph in the exercise figure is y = a \cdot b^x. Read a and b from the labeled points and write the rule for A and for B.
Day 2: contexts
- In a second pond, duckweed covers 2 square meters when first measured and triples every week (made-up numbers). The pond’s area is 162 square meters. Write the rule for the cover C after t weeks, find when the pond is covered, and give a sensible domain and range.
- A second tracer has strength 64 units at t = 0 and halves every hour (made-up numbers). Write the rule, find the strength at t = 0, 3 and 6, and give the range of the graph drawn for 0 \le t \le 8.
- For the tracer in Example 6, at which whole hour is the strength first below 0.1 units?
Day 2: reasoning
- Describe the graph of y = 5 \cdot 1^x, and explain why b = 1 is left out of the family.
- Find (-2)^x for x = 1, 2, 3, 4, and explain why y = (-2)^x has no smooth graph.
- Oswin says the graphs of y = 2^x and y = 3^x never meet. Say where they meet and which is higher on each side, and explain why.
Answers
- \displaystyle \frac{1}{16}, \frac{1}{4}, 1, 4, 16; the y-intercept is (0, 1). For instance \displaystyle 4^{-2} = \frac{1}{4^2} = \frac{1}{16}.
- \displaystyle 16, 4, 1, \frac{1}{4}, \frac{1}{16}, the table of exercise 1 read backward, since \displaystyle \left(\frac{1}{4}\right)^x = 4^{-x}.
- \displaystyle \left(-4, \frac{1}{16}\right), (0, 1) and (5, 32), since \displaystyle 2^{-4} = \frac{1}{16}, 2^0 = 1 and 2^5 = 32.
- Domain all real numbers; range y > 0; y-intercept (0, 1) as 5^0 = 1; asymptote y = 0, on the left.
- The graph falls, leveling toward 0 on the right and growing without bound on the left: as x \to \infty, y \to 0; as x \to -\infty, y \to \infty.
- (-3, 8), the same height on the other side of the y-axis. Check: \displaystyle \left(\frac{1}{2}\right)^{-3} = 2^3 = 8.
- (-2, -4) is not on it: \displaystyle 2^{-2} = \frac{1}{4}, and no output of 2^x is negative. \displaystyle \left(-2, \frac{1}{4}\right) is on it.
- Between 6 and 7: 2^6 = 64 is below 100 and 2^7 = 128 is above, and the graph rises without a break.
- Between -7 and -6: \displaystyle 2^{-6} = \frac{1}{64} \approx 0.0156 is above 0.01 and \displaystyle 2^{-7} = \frac{1}{128} \approx 0.0078 is below.
- Going left, each step of 1 halves the output. Half of a positive number is positive, so starting from 2^0 = 1 every output stays above 0: the curve keeps getting closer to the axis but never reaches it. The x-axis is an asymptote, not a crossing.
- \displaystyle \left(\frac{1}{2}\right)^x = \frac{1^x}{2^x} = \frac{1}{2^x} = 2^{-x} for every x, so the two rules give the same output everywhere. The output at x is 2^x’s output at -x, so each point (x, y) of y = 2^x appears as (-x, y): the reflection in the y-axis.
- \displaystyle \frac{3}{25}, \frac{3}{5}, 3, 15, 75; y-intercept (0, 3); asymptote y = 0, on the left. For instance \displaystyle 3 \cdot 5^{-2} = \frac{3}{25}.
- 16, 8, 4, 2, 1; the graph falls, since \displaystyle 0 < \frac{1}{2} < 1. As x \to \infty, y \to 0; as x \to -\infty, y \to \infty.
- (a) rises, (0, 7), since 1.2 > 1; (b) rises, (0, 0.5); (c) falls, (0, 10), since 0 < 0.8 < 1.
- At x = 3, y = 4^x, with 64 against 8 and 3.375. At x = -3, y = 1.5^x, with \displaystyle \frac{8}{27} \approx 0.296 against \displaystyle \frac{1}{8} and \displaystyle \frac{1}{64}.
- A: a = 4 from (0, 4), and \displaystyle b = \frac{10}{4} = 2.5, so y = 4 \cdot 2.5^x. B: a = 9 and \displaystyle b = \frac{3}{9} = \frac{1}{3}, so \displaystyle y = 9 \cdot \left(\frac{1}{3}\right)^x; check (2, 1): \displaystyle 9 \cdot \frac{1}{9} = 1.
- C = 2 \cdot 3^t. Covered at t = 4, since 2 \cdot 3^4 = 162. Domain 0 \le t \le 4, range 2 \le C \le 162.
- \displaystyle S = 64 \cdot \left(\frac{1}{2}\right)^t; 64, 8 and 1 units. The graph falls from 64 to \displaystyle 64 \cdot \left(\frac{1}{2}\right)^8 = \frac{1}{4}, so the range is \displaystyle \frac{1}{4} \le S \le 64.
- At t = 10: \displaystyle 80 \cdot \left(\frac{1}{2}\right)^9 \approx 0.156 is still above 0.1, and \displaystyle 80 \cdot \left(\frac{1}{2}\right)^{10} \approx 0.078 is below.
- Since 1^x = 1, the rule is y = 5 for every x: a horizontal line. Nothing is multiplied up or down from step to step, so it is neither growth nor decay, and it has none of the family’s features (no asymptote, a range of one number).
- -2, 4, -8, 16. The outputs jump from one side of the x-axis to the other, and at \displaystyle x = \frac{1}{2} the value would be a real number whose square is -2, which does not exist. The points cannot be joined by one unbroken curve.
- They meet only at (0, 1), since 2^0 = 3^0 = 1. Compare by dividing: \displaystyle \frac{3^x}{2^x} = \left(\frac{3}{2}\right)^x, a rising exponential through (0, 1). So the ratio is above 1 for x > 0, where y = 3^x is higher, and below 1 for x < 0, where y = 2^x is higher; it equals 1 only at x = 0.