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B. Exponential Functions · Exponential Functions

Finding the Equation of an Exponential Function

B2.5About 70 minutes
Learning goals
  • I can find c from the asymptote, a from the y-intercept, and b from another point.
  • I can find y = a(bˣ) from two points by dividing to eliminate a.
  • I can write an equation from a list of properties such as range, intercept and increasing/decreasing.
  • I can check that my equation matches every given point and the direction of the graph.

Words to know

y = a(bˣ) + c
A useful form for an exponential function: c is the asymptote, a + c is the y-intercept, and b is the base (the growth or decay factor).
initial value
The output when x = 0. For y = a(bˣ) + c it is a + c; for y = a(bˣ) it is a.
growth or decay factor
The base b: each step of 1 in x multiplies the distance from the asymptote by b.

Working backwards

So far you started with an equation and drew a graph. Now reverse it: given a graph or a list of properties, find the equation. This is what scientists do when they fit a model to data. Most exponential functions in this course can be written in the form

y = a(bˣ) + c

(the form y = a(b^(k(x − d))) + c can always be rewritten this way, as Lesson 10 showed). Three unknowns means you need three pieces of information, and the asymptote is usually the easiest one to read.

A step-by-step plan

  1. c = the horizontal asymptote (y = c).
  2. y-intercept: substitute x = 0. Since b⁰ = 1, the y-intercept is a + c, so a = (y-intercept) − c.
  3. One more point (x, y): substitute to find b. Remember b > 0.
  4. Check with a point you did not use, and check that the graph's direction (increasing or decreasing, above or below the asymptote) matches the signs of a and b.

Why the plan works

The distance from the curve to its asymptote is a(bˣ). At x = 0 that distance is a. Every step of 1 in x multiplies the distance by b. So once you know the asymptote and the starting distance, one more point tells you the multiplying factor. If a > 0 the graph is above the asymptote; if a < 0 it is below. If b > 1 the distance grows to the right; if 0 < b < 1 it shrinks to the right.

xy−2−101234−10123456789101112(0, 7)(1, 4.5)(−1, 12)y = 2
The graph has asymptote y = 2 and passes through (0, 7) and (1, 4.5). The distance from the asymptote goes 10 → 5 → 2.5: it is multiplied by 1/2 at each step.
Worked example

Example 1: Equation from a graph

Find the equation of the exponential function in the graph above: asymptote y = 2, through (0, 7) and (1, 4.5).

  1. c = 2 (the asymptote).
  2. y-intercept 7 = a + 2, so a = 5.
  3. Point (1, 4.5): 4.5 = 5b¹ + 2, so 5b = 2.5 and b = 0.5.
  4. Check with (−1, 12): 5(0.5)⁻¹ + 2 = 5(2) + 2 = 12 ✓. The graph decreases and is above the asymptote: a > 0 and 0 < b < 1 ✓.

y = 5(0.5)ˣ + 2

xy−3−2−1012−24−20−16−12−8−40(0, −6)(1, −10)(2, −22)asymptote y = −4
This graph lies below its asymptote y = −4 and passes through (0, −6) and (2, −22), so a must be negative.
Worked example

Example 2: A graph below its asymptote

Find the equation of the graph above: asymptote y = −4, through (0, −6) and (2, −22).

  1. c = −4.
  2. y-intercept: −6 = a − 4, so a = −2.
  3. Point (2, −22): −22 = −2b² − 4, so −2b² = −18, b² = 9 and b = 3 (b must be positive).
  4. Check with (1, −10): −2(3) − 4 = −10 ✓.

y = −2(3ˣ) − 4

Worked example

Example 3: Two points and no vertical shift

An exponential function y = a(bˣ) (asymptote y = 0) passes through (1, 10) and (3, 250). Find its equation.

  1. Substitute both points: ab = 10 and ab³ = 250.
  2. Divide the second equation by the first: ab³ ÷ ab = b² = 25, so b = 5.
  3. Then a = 10 ÷ 5 = 2.
  4. Check: 2(5³) = 250 ✓.

y = 2(5ˣ)

Worked example

Example 4: Equation from properties

Find an exponential function with range {y ∈ ℝ | y > −1}, that is increasing, has y-intercept 2, and has f(1) = 5.

  1. Range y > −1: asymptote c = −1, and the graph is above it, so a > 0.
  2. y-intercept: a + (−1) = 2, so a = 3.
  3. f(1) = 5: 3b − 1 = 5, so b = 2. Increasing needs b > 1 ✓.
  4. Equation: f(x) = 3(2ˣ) − 1.

f(x) = 3(2ˣ) − 1

Worked example

Example 5: Using the transformation form

The graph of an exponential function is congruent to y = 2ˣ (same shape and orientation), has asymptote y = 1, and passes through (3, 2) and (4, 3). Write its equation in the form y = 2^(x − d) + c, then in the form y = a(2ˣ) + c.

  1. Same shape and orientation as y = 2ˣ means a = 1 and k = 1. The asymptote gives c = 1.
  2. On y = 2ˣ, the point (0, 1) is one unit above the asymptote. Here (3, 2) is one unit above y = 1. So the graph moved right 3: d = 3.
  3. Equation: y = 2^(x − 3) + 1. Check (4, 3): 2¹ + 1 = 3 ✓.
  4. Rewrite: 2^(x − 3) = 2ˣ × 2⁻³ = (1/8)(2ˣ). So y = (1/8)(2ˣ) + 1.

y = 2^(x − 3) + 1, which is the same as y = (1/8)(2ˣ) + 1.

Just for fun

Finding an exponential equation is a detective story. The asymptote has an alibi ("I was at y = 2 the whole time"), the y-intercept leaves a fingerprint, and one more point closes the case.

Case closed: y = 5(0.5)ˣ + 2. The base did it.

Hands-on activity

Design and decode

You need: grid paper, a pencil, a ruler, a calculator and a partner.

  1. Secretly choose a function y = a(bˣ) + c with a whole-number a between −4 and 4 (not 0), b = 2, 3 or 1/2, and a whole-number c between −3 and 3.
  2. On grid paper draw the asymptote as a dashed line and plot three points: x = 0, x = 1 and x = 2 (or x = −1 if b = 1/2, to keep numbers small). Draw a smooth curve.
  3. Swap graphs with your partner. Each of you finds the other's equation using the plan: c from the asymptote, a from the y-intercept, b from a third point. Show your work.
  4. Check by substituting the remaining point.

What to look for: for example, a = −3, b = 2, c = 1 gives the points (0, −2), (1, −5), (2, −11) below the asymptote y = 1. Decoding: c = 1; −2 = a + 1 gives a = −3; −5 = −3b + 1 gives b = 2. Check (2, −11): −3(4) + 1 = −11 ✓.

Your turn

1

An exponential graph has asymptote y = 0 and passes through (0, 4) and (1, 12). Find y = a(bˣ).

2

Asymptote y = 5, y-intercept 8, passes through (1, 11). Find y = a(bˣ) + c.

3

Asymptote y = −2, y-intercept 14, passes through (2, 2). Find the equation.

4

y = a(bˣ) passes through (2, 18) and (4, 162). Find a and b.

5

Range {y ∈ ℝ | y < 3}, y-intercept 1, and f(1) = −1. Find f(x) = a(bˣ) + c.

6

A graph has the shape of y = 3ˣ, asymptote y = −1, and passes through (2, 0). Write it in the form y = 3^(x − d) + c.

More practice: warm-up

1

What does c tell you in y = a(bˣ) + c?

2

For y = a(bˣ) + c, what is the y-intercept?

3

A graph is above its asymptote. Is a positive or negative?

4

y = a(bˣ) has y-intercept 7. What is a?

5

A decreasing graph above its asymptote: is b greater than 1 or between 0 and 1?

More practice: on your own

1

Asymptote y = 10, y-intercept 4, passes through (1, 7). Find the equation. Is it increasing or decreasing?

2

y = a(bˣ) passes through (−1, 2) and (1, 18). Find the equation.

3

Asymptote y = 0, through (0, 200) and (3, 25). Find the equation.

4

A function has domain ℝ, range y > 4, is decreasing, has y-intercept 7 and passes through (−2, 31). Find it.

5

Write y = 2^(x + 2) − 5 in the form y = a(bˣ) + c and state the y-intercept.

Challenge

1

y = a(bˣ) + 3 passes through (1, 5) and (2, 11). Find a and b.

2

An exponential function has asymptote y = 2 and passes through (0, 5), (1, 8) and (2, 14). Do these points fit a function of the form y = a(bˣ) + 2? Explain.

3

Explain why two points and the asymptote are enough to find y = a(bˣ) + c, but two points alone are not.

Check your understanding

Quiz

1.

A graph has asymptote y = −3. In y = a(bˣ) + c, c is:

2.

y = a(bˣ) + 2 has y-intercept 9. What is a?

3.

y = 3(bˣ) passes through (2, 48). What is b?

4.

Which equation has asymptote y = 1, y-intercept 5, and passes through (1, 3)?

5.

A graph lies below its asymptote y = 6 and increases. Which could be its equation?

6.

y = a(bˣ) passes through (1, 6) and (3, 54). What is the equation?

For parents and teachers

Big idea: Expectation B2.5: represent an exponential function with an equation, given its graph or its properties.

How to help: Ask your teen to say the plan out loud: "asymptote gives c, y-intercept gives a, another point gives b, then check." Always ask for the check: it catches most errors.

Common difficulties:

  • Using the y-intercept as a when the graph has been shifted: a = (y-intercept) − c.
  • Taking a negative root for b: b must be positive (b² = 9 gives b = 3 only).
  • Mixing up the sign of a with the direction of the graph. Above or below the asymptote decides the sign of a; growing or shrinking distance decides whether b > 1.
  • Forgetting that y = 2^(x − 3) + 1 and y = (1/8)(2ˣ) + 1 are the same function written two ways.