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B. Exponential Functions · Exponential Functions

Transforming Exponential Graphs II: Horizontal Stretches, Reflections and All Four Parameters

B2.2 B2.3About 75 minutes
Learning goals
  • I can describe how k in y = f(kx) stretches, compresses or reflects the graph of f(x) = bˣ.
  • I can factor the exponent to find k and d correctly.
  • I can sketch y = af(k(x − d)) + c using the mapping rule (x, y) → (x/k + d, ay + c).
  • I can state the domain, range, asymptote and intercepts of a transformed exponential function.

Words to know

horizontal stretch / compression
In y = f(kx), every x-coordinate is multiplied by 1/k: a compression toward the y-axis if |k| > 1, a stretch away from it if 0 < |k| < 1.
reflection in the y-axis
In y = f(−x) every x-coordinate changes sign, so the graph flips over the y-axis.
mapping rule
A rule that tells where every point goes: for y = af(k(x − d)) + c it is (x, y) → (x/k + d, ay + c).
factored form of the exponent
Writing kx − kd as k(x − d), so that d can be read correctly.

The last parameter: k

The parameter k acts on x, inside the function. In y = 2^(kx), the input is multiplied by k before the base is raised to it. That changes the graph horizontally, and in the opposite way to what you might expect: multiplying x by 2 makes the graph narrower, not wider.

Investigation: y = 2^(kx)

Graph y = 2ˣ, y = 2^(2x) and y = 2^(x/2) (with technology or tables).

  • y = 2ˣ reaches 4 at x = 2. y = 2^(2x) reaches 4 when 2x = 2, that is at x = 1. Every point is half as far from the y-axis: a horizontal compression by a factor of 1/2.
  • y = 2^(x/2) reaches 4 when x/2 = 2, that is at x = 4: a horizontal stretch by a factor of 2.
  • y = 2^(−x): the point (2, 4) moves to (−2, 4). Negative k reflects the graph in the y-axis. Notice 2^(−x) = (1/2)ˣ: you saw this reflection in Lesson 1.

So in y = f(kx), each x-coordinate is divided by k (multiplied by 1/k). The y-coordinates and the asymptote do not change.

xy−3−2−10123012345678× 1/2(2, 4)(1, 4)(2, 2)y = 2^(2x)y = 2ˣy = 2^(x/2)
y = 2^(2x) (long dashes) is y = 2ˣ compressed horizontally by 1/2: (2, 4) → (1, 4). y = 2^(x/2) (dotted) is stretched horizontally by 2: (2, 2) on it comes from (1, 2) on y = 2ˣ.

The full mapping rule

For y = af(k(x − d)) + c with f(x) = bˣ:

(x, y) → (x/k + d, ay + c)

  • x-side (inside): divide by k (horizontal stretch/compression, reflect in the y-axis if k < 0), then add d (translate).
  • y-side (outside): multiply by a (vertical stretch/compression, reflect in the x-axis if a < 0), then add c (translate).
  • Asymptote: y = c. Domain: {x ∈ ℝ}. Range: y > c if a > 0; y < c if a < 0.

Stretches and reflections come before translations.

Factor first!

In y = 2^(3x − 6), the graph does not move right 6. Factor the exponent: 3x − 6 = 3(x − 2). So k = 3 and d = 2: compress horizontally by 1/3, then move right 2. Always write the exponent as k(x − d) before reading d.

Worked example

Example 1: A horizontal compression

Sketch y = 2^(2x) using the key points of y = 2ˣ.

  1. k = 2, so (x, y) → (x/2, y).
  2. (−1, 1/2) → (−0.5, 1/2); (0, 1) → (0, 1); (1, 2) → (0.5, 2); (2, 4) → (1, 4).
  3. The y-intercept (0, 1) does not move: 0 ÷ 2 = 0. The asymptote stays y = 0.
  4. Notice: 2^(2x) = (2²)ˣ = 4ˣ. The compressed graph is just y = 4ˣ. (Lesson 10 explores this.)

A steeper curve through (−0.5, 1/2), (0, 1), (0.5, 2), (1, 4).

Worked example

Example 2: Reflection in the y-axis with a stretch

Describe and sketch y = 3^(−x/2).

  1. k = −1/2. 1/k = −2: stretch horizontally by a factor of 2 and reflect in the y-axis.
  2. Mapping: (x, y) → (−2x, y).
  3. Key points of y = 3ˣ: (−1, 1/3) → (2, 1/3); (0, 1) → (0, 1); (1, 3) → (−2, 3); (2, 9) → (−4, 9).
  4. The graph now decreases. Asymptote y = 0, approached on the right.

Horizontal stretch by 2 and reflection in the y-axis; decreasing curve through (−4, 9), (−2, 3), (0, 1), (2, 1/3).

Worked example

Example 3: All four parameters

Sketch y = −2(2^(3x − 6)) + 1. State the domain, range, asymptote and any intercepts.

  1. Factor the exponent: 3x − 6 = 3(x − 2). So a = −2, k = 3, d = 2, c = 1.
  2. Transformations: horizontal compression by 1/3; translate right 2; reflect in the x-axis and stretch vertically by 2; translate up 1.
  3. Mapping: (x, y) → (x/3 + 2, −2y + 1).
  4. (−1, 1/2) → (−1/3 + 2, −1 + 1) = (5/3, 0); (0, 1) → (2, −1); (1, 2) → (7/3, −3); (2, 4) → (8/3, −7).
  5. Asymptote y = 1, graph below it (a < 0). Domain {x ∈ ℝ}; range {y ∈ ℝ | y < 1}.
  6. x-intercept: (5/3, 0) (a key point landed on the axis). y-intercept: x = 0 gives −2(2⁻⁶) + 1 = −2/64 + 1 = 31/32.

Points (5/3, 0), (2, −1), (7/3, −3), (8/3, −7); asymptote y = 1; range y < 1; x-intercept 5/3, y-intercept 31/32.

xy01234−8−7−6−5−4−3−2−1012(5/3, 0)(2, −1)(7/3, −3)(8/3, −7)asymptote y = 1y = −2(2^(3x − 6)) + 1
y = −2(2^(3x − 6)) + 1. The key points (5/3, 0), (2, −1), (7/3, −3) and (8/3, −7) come from (−1, 1/2), (0, 1), (1, 2) and (2, 4) on y = 2ˣ. The curve approaches the asymptote y = 1 from below on the left.
Worked example

Example 4: Describing from the equation

Describe the transformations of f(x) = 3ˣ that give y = 5(3^(−2x + 4)) + 2, and state the range.

  1. Factor: −2x + 4 = −2(x − 2). So k = −2, d = 2, a = 5, c = 2.
  2. Horizontal compression by a factor of 1/2 and reflection in the y-axis (k = −2).
  3. Vertical stretch by a factor of 5 (a = 5).
  4. Translation right 2 and up 2 (d = 2, c = 2).
  5. Range: a > 0, so {y ∈ ℝ | y > 2}.

Compress horizontally by 1/2, reflect in the y-axis, stretch vertically by 5, translate right 2 and up 2; range y > 2.

Worked example

Example 5: Equation from a description

The graph of f(x) = 4ˣ is compressed vertically by a factor of 1/2, reflected in the x-axis, stretched horizontally by a factor of 3, and translated 1 left and 2 down. Write the equation.

  1. a: compression by 1/2 and reflection in the x-axis → a = −1/2.
  2. k: horizontal stretch by 3 → k = 1/3.
  3. d = −1 (left 1), c = −2 (down 2).

y = −(1/2)(4^((1/3)(x + 1))) − 2

Just for fun

Transformations are like getting dressed on a snowy morning: socks before boots. Stretch and reflect first, then translate.

Do it the other way around and you will still get out the door, but your socks will be on the outside, and your graph will be in the wrong place.

Hands-on activity

Transformation dice challenge

You need: two number cubes (dice), paper, a pencil and a calculator. Desmos (free on a phone or computer) is helpful for checking but not required.

  1. Roll one die for d: 1 → −2, 2 → −1, 3 → 0, 4 → 1, 5 → 2, 6 → 3. Roll the other for c using the same table.
  2. Flip a coin (or roll: even/odd) for a: heads → a = 2, tails → a = −1. Flip again for k: heads → k = 2, tails → k = −1.
  3. Write the equation y = a(2^(k(x − d))) + c. Write the mapping rule, transform the four key points of y = 2ˣ, and sketch the graph with its asymptote.
  4. Check one point by substituting its x-value into your equation on the calculator. Check the whole sketch in Desmos if you can.
  5. Repeat three times. Have a family member pick a point on your sketch and ask you to prove it is on the graph.

What to look for: for example, d = 1, c = −2, a = −1, k = 2 gives y = −2^(2(x − 1)) − 2, mapping (x, y) → (x/2 + 1, −y − 2). The key points become (0.5, −2.5), (1, −3), (1.5, −4), (2, −6), the asymptote is y = −2 and the range is y < −2. Substituting x = 1.5: −2^(1) − 2 = −4 ✓.

Your turn

1

Describe the transformation: y = 2^(3x).

2

Describe the transformation: y = 5^(x/4).

3

Describe the transformation: y = 3^(−x).

4

Factor the exponent and state k and d: y = 2^(4x + 8).

5

Write the mapping rule for y = 3(2^(2(x − 1))) − 4.

6

Find the image of (2, 4) on y = 2ˣ under y = −(2^(0.5(x + 3))) + 6.

7

State the domain, range and asymptote of y = −4(3^(2x − 1)) − 5.

More practice: warm-up

1

Which parameter causes horizontal stretches and compressions?

2

y = 2^(x/3): stretch or compression? By what factor?

3

Under y = 2^(−x), where does (1, 2) go?

4

Does k change the asymptote?

5

Write 2^(5x − 10) with the exponent factored.

More practice: on your own

1

Sketch y = 2^(2(x + 1)) + 3 using the key points of y = 2ˣ.

2

Describe all transformations of f(x) = 2ˣ for y = (1/3)(2^(−(x − 4))) + 1.

3

Write the equation: y = 3ˣ is stretched horizontally by 2, reflected in the x-axis, and translated 5 right.

4

For y = 2^(6 − 2x), find k and d. (Hint: factor out −2.)

5

Find the y-intercept of y = −2(2^(3x − 6)) + 1 and explain why it is very close to the asymptote.

Challenge

1

Show that y = 2^(2x) and y = 4ˣ are the same, and explain why a horizontal compression by 1/2 of y = 2ˣ looks like a different base.

2

A graph has asymptote y = 3, lies above it, and passes through (2, 4) and (3, 5). It has the form y = 2^(k(x − d)) + 3. Find k and d.

3

Explain why y = 2^(−x) + 1 and y = (1/2)ˣ + 1 have identical graphs, and give the range.

Check your understanding

Quiz

1.

y = 2^(4x) is y = 2ˣ:

2.

For y = 2^(2x − 8), what is the horizontal translation?

3.

What is the mapping rule for y = 3(2^(−x)) + 1?

4.

Which equation is the same as y = 5^(−x)?

5.

What is the range of y = −2(3^(2(x − 1))) + 4?

6.

Under y = 2^(0.5x), the point (3, 8) on y = 2ˣ maps to:

For parents and teachers

Big idea: Expectations B2.2 and B2.3, completed: the role of k, and sketching y = af(k(x − d)) + c with all four parameters, stating domain and range.

How to help: Ask your teen to factor the exponent before anything else, then to say the mapping rule out loud. A quick check: pick an x, substitute it in the equation on a calculator, and see if the point is on their sketch.

Common difficulties:

  • Multiplying x-coordinates by k instead of dividing.
  • Reading d before factoring: 2^(3x − 6) moves right 2, not 6.
  • Doing translations before stretches.
  • Mixing up the two reflections: a < 0 flips over the x-axis; k < 0 flips over the y-axis.