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Lesson 16: Graphing Transformations of Exponential Graphing Transformations of Exponential
Lesson 16: Graphing Transformations of Exponential Functions
Functions
Learning Outcome(s)
Learning Outcome(s): At the end of the lesson, the learner is able to graph exponential functions.
Lesson Outline:
Lesson Outline:
1. Vertical and horizontal reflection 2. Stretching and shrinking 3. Vertical and horizontal shifts Reflecting Graphs
Reflecting Graphs Example 1.
Example 1. Use the graph of y = 2x to graph the functions y = –2x and y = 2 –x. Solution. Some y-values are shown on the following table.
x –3 –2 –1 0 1 2 3
y = 2x 0.125 0.25 0.5 1 2 4 8
y = –2x –0.125 –0.25 –0.5 –1 –2 –4 –8
y = 2 –x 8 4 2 1 0.5 0.25 0.125
The y-coordinate of each point on the graph of y = –2x the negative of the y-coordinate of the graph of y = 2x. Thus, the graph of y = –2x is the reflection of the graph of y = 2x about the x-axis.
The value of y = 2 –x at x is the same as the value of y = 2xat –x. Thus, the graph of y = 2 –x is the reflection of the graph of y = 2x about the y-axis.
The corresponding graphs are shown below.
D E P E D C O P Y
The results in Example 1can be generalized as follows:
Reflection Reflection
The graph of y = –f(x) is the reflection about the x-axis of the graph of y = f(x).
The graph of y = f( –x) is the reflection about the y-axis of the graph of y = f(x).
Example 2.
Example 2.Use the graph of y = 2x to graph the functions y = 3(2x) and y = 0.4(2x).
Solution. Some y-values are shown on the following table.
x –3 –2 –1 0 1 2 3
y = 2x 0.125 0.25 0.5 1 2 4 8
y = 3(2x) 0.375 0.75 1.5 3 6 12 24
y = 0.4(2x) 0.05 0.1 0.2 0.4 0.8 1.6 3.2
The y-coordinate of each point on the graph of y = 3(2x) is 3 times the y-coordinate of each point on y = 2x. Similarly, the y-coordinate of each point on the graph of y = 0.4(2x) is 0.4 times the y-coordinate of each point on y = 2x.
The graphs of these functions are shown below.
Observations.
Observations.
1. The domain for all three graphs is the set of all real numbers.
2. The y-intercepts were also multiplied correspondingly. The y-intercept of y = 3(2x) is 3, and the y-intercept of y = 0.4(2x) is 0.4.
3. All three graphs have the same horizontal asymptote: y = 0.
4. The range of all three graphs is the set of all y > 0.
D E P E D C O P Y
The results of Example 2 can be generalized as follows.
Vertical Stretching or Shrinking Vertical Stretching or Shrinking
Let c be a positive constant. The graph of y = cf(x) can be obtained from the graph of y = f(x) by multiplying each y-coordinate by c. The effect is a vertical stretching (if c >
1) or shrinking (if c < 1) of the graph of y = f(x).
Example 3.
Example 3. Use the graph of y = 2x to graph y = 2x – 3 and y = 2x + 1.
Solution .. Some y-values are shown on the following table.
x –3 –2 –1 0 1 2 3
y = 2x 0.125 0.25 0.5 1 2 4 8
y = 2x – 3 –2.875 –2.75 –2.5 –2 –1 1 5
y = 2x + 1 1.125 1.25 1.5 2 3 5 9
The graphs of these functions are shown below.
Observations.
Observations.
1. The domain for all three graphs is the set of all real numbers.
2. The y-intercepts and horizontal asymptotes were also vertically translated from the y-intercept and horizontal asymptote of y = 2x.
3. The horizontal asymptote of y = 2x is y = 0. Shift this 1 unit up to get the horizontal asymptote of y = 2x + 1 which is y = 1, and 3 units down to get the horizontal asymptote of y = 2x – 3 which is y = – 3.
4. The range of y = 2x + 1 is all y > 1, and the range of y = 2x – 3 is all y > –3.
D E P E D C O P Y
The results of Example 3 can be generalized as follows.
Vertical Shifts Vertical Shifts
Let k be a real number. The graph of y = f(x) + k is a vertical shift of k units up (if k >
0) or k units down (if k < 0) of the graph of y = f(x).
Example 4.
Example 4. Use the graph of y = 2x to graph y = 2x –2and y = 2x+4 Solution .. Some y-values are shown on the following table.
x –3 –2 –1 0 1 2 3
y = 2x 0.125 0.25 0.5 1 2 4 8
y = 2x –2 0.031 0.063 0.125 0.25 0.5 1 2
y = 2x+4 2 4 8 16 32 64 128
The graphs of these functions are shown below.
Observations.
Observations. 1. The domain for all three graphs is the set of all real numbers.
2. The y-intercepts changed. To find them, substitute x = 0 in the function. Thus, the y-intercept of y = 2x+4 is 24 = 16 and the y-intercept of y = 2x –2 is 2 –2 = .25.
3. The horizontal asymptotes of all three graphs are the same (y = 0). Translating agraph horizontally does not change the horizontal asymptote.
4. The range of all three graphs is the set of all y > 0.
D E P E D C O P Y
The results of Example 4 can be generalized as follows.
Horizontal Shifts Horizontal Shifts
Let k be a real number. The graph of y = f(x – k) is a horizontal shift of k units to the right (if k > 0) or k units to the left (if k < 0) of the graph of y = f(x).
Solved Examples Solved Examples
1. Sketch the graph of F(x) = 3x+1 – 2, then state the domain, range, y-intercept, and horizontal asymptote.
Solution.
Transformation: The base function f(x) = 3x will be shifted 1 unit to the left and 2 units down.
Steps in Graph Sketching:
Step 1: Base function: f(x) = 3x ; y-intercept: (0,1) ; horizontal asymptote: y = 0 Step 2: The graph of F(x) is found by shifting the graph of the function f left one unit
and down two units.
Step 3: The y-intercept of f(x) (0,1) will shift to the left by one unit and down two units towards ( –1, –1). This is not the y-intercept of F(x).
Step 4: The horizontal asymptote will be shifted down two units, which is y = –2.
Step 5: Find additional points on the graph; F(0) = 30+1 – 2 = 1 and F(1) =31+1 – 2 = 7.
[(0,1) and (1,7)]
Step 6: Connect the points using a smooth curve.
Domain: All real numbers Range: ( –2, ∞)
y-intercept: (0,1)
Horizontal Asymptote: y = –2
2. Sketch the graph of G(x) = 4
, then state the domain, range, y-intercept, and horizontal asymptote.
D E P E D C O P Y
Solution.
Transformation: The base function g(x) = (1/2)x will be stretched 4 units (that is, every y-value will be multiplied by 4), then will be shifted 1 unit upward.
Steps in Graph Sketching:
Step 1: Base function: g(x) =
; y-intercept: (0,1) ; horizontal asymptote: y = 0 Step 2: The graph of G(x) is obtained by stretching the graph of g by four units thenshifting the graph upward by one unit.
Step 3: Since the graph will be stretched by 4 units, the y-intercept of g(x) (0,1) will be at (0,4), then will be shifted again by 1 unit upward to get (0, 5). This is the y-intercept of G(x).
Step 4: The horizontal asymptote will be shifted 1 unit upward, which is y = 1.
Step 5: Find additional points in the graph: G(-1) = 4(1/2)-1 + 1 = 9 and G(3) = 4(1/2)3 + 1 = 3/2. [( –1,1/2) and (3, 3/2)]
Step 6: Connect the points using a smooth curve.
Domain: All real numbers Range: (1, ∞)
y-intercept: (0,5)
Horizontal Asymptote: y = 1
Lesson 16 Supplementary Exercises Lesson 16 Supplementary Exercises
In Exercises 1-3, (a)(a) use transformations to describe how the graph is related to its base exponential function y = bx, (b)(b) sketch the graph, (c)(c) identify its domain, range, y-intercept, and horizontal asymptote.
1. F(x) = 2·3x 2. G(x) = (1/4) x+14 – 3. H(x) = –2(3x –1) 4. Find an exponential function of the form f(x) = abx + c such that the y-intercept is
–5, the horizontal asymptote is y = –10, and f(2) = 35.
5. Give the range of the function y = 3x for –10 x 10.
D E P E D C O P Y
Lessons 12
Lessons 12 – – 16 Topic Test 1 16 Topic Test 1
1. Solve for x. [10][10]
a. 3-x = 27x+2 b.
2. Solve the inequality 5x>
. [10][10]3. The population of a certain city doubles every 50 years. [15][15]
(a) Give an exponential model for this situation.
(b) By what factor does the population increase after 30 years?
(c) If the city’s population is currently 100,000, how long will it take for the population to exceed 400,000?
4. Graph the following functions. Label all intercepts and asymptotes. Indicate the
domain and range. [15][15]
a. f(x) = 3+ 1 x-2 b. h(x) = (0.1) x+2 – 1
Lessons 12
Lessons 12 – – 16 Topic Test 2 16 Topic Test 2
1. Solve for x in the following equations. [10][10]
a.
b. 81-2x = 162x-12. Solve the inequality
[10][10]3. Maine decides to participate in an investment that yields 3.75% interest annually.
If she invests ₱12,500, how much will her investment be after 5 years? [10][10]
4. The population of a certain bacteria colony is modeled by the function P(x) = 500e0.05t, where P(x) is the bacteria population after t minutes. Find the
bacteria population after half an hour. [10][10]
5. Graph the function f(x) = 2(1/2)x-1 . Label all intercepts and asymptotes. Indicate
the domain and range. [10][10]