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Transformations of Exponential Functions

A free Precalculus lesson from the “Exponential and Logarithmic Functions” unit, with a worked example and practice problems including step-by-step solutions.

Exponential transformations change the asymptote, direction, and scale without changing the repeated-multiplier idea. This lesson is part of Precalculus: Advanced Functions, so the emphasis is on interpreting behavior, choosing the right representation, and explaining the result clearly rather than memorizing isolated algebra moves.

What you'll learn

Why it matters: Exponential and logarithmic models describe growth, decay, sound, pH, finance, and scientific scales.

Worked example

Problem. Compared with y = 2^x, y = 2^x + 3 has horizontal asymptote:

  1. Worked Example: First identify exactly what the question is asking: Compared with y = 2^x, y = 2^x + 3 has horizontal asymptote:
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. The parent exponential has asymptote y = 0.
  4. Adding 3 shifts the graph up 3.
  5. The asymptote becomes y = 3.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: y = 3

Practice problems

1. Compared with y = 2^x, y = 2^x + 3 has horizontal asymptote:

Choices: y = 3 · y = 0 · x = 3 · y = -3

Show solution
  1. Warm-up: First identify exactly what the question is asking: Compared with y = 2^x, y = 2^x + 3 has horizontal asymptote:
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. The parent exponential has asymptote y = 0.
  4. Adding 3 shifts the graph up 3.
  5. The asymptote becomes y = 3.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: y = 3

2. Compared with y = 2^x, y = 2^x + 4 has horizontal asymptote:

Choices: y = 4 · y = 0 · x = 4 · y = -4

Show solution
  1. Warm-up: First identify exactly what the question is asking: Compared with y = 2^x, y = 2^x + 4 has horizontal asymptote:
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. The parent exponential has asymptote y = 0.
  4. Adding 4 shifts the graph up 4.
  5. The asymptote becomes y = 4.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: y = 4

3. Compared with y = 2^x, y = 2^x + 5 has horizontal asymptote:

Choices: y = 5 · y = 0 · x = 5 · y = -5

Show solution
  1. Core Practice: First identify exactly what the question is asking: Compared with y = 2^x, y = 2^x + 5 has horizontal asymptote:
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. The parent exponential has asymptote y = 0.
  4. Adding 5 shifts the graph up 5.
  5. The asymptote becomes y = 5.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: y = 5

4. Compared with y = 2^x, y = -2^x is:

Choices: reflected across the x-axis · shifted right 2 · reflected across the y-axis · unchanged

Show solution
  1. Core Practice: First identify exactly what the question is asking: Compared with y = 2^x, y = -2^x is:
  2. For signed numbers, track both distance from zero and direction so the sign of the answer makes sense.
  3. The negative sign is outside the exponential.
  4. It changes output signs.
  5. That reflects over the x-axis.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: reflected across the x-axis

5. Compared with y = 2^x, y = -2^x is: (variation 2)

Choices: reflected across the x-axis · shifted right 2 · reflected across the y-axis · unchanged

Show solution
  1. Core Practice: First identify exactly what the question is asking: Compared with y = 2^x, y = -2^x is:
  2. For signed numbers, track both distance from zero and direction so the sign of the answer makes sense.
  3. The negative sign is outside the exponential.
  4. It changes output signs.
  5. That reflects over the x-axis.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: reflected across the x-axis

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