1. 6.1 VERTICAL AND HORIZONTAL SHIFTS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

2. Vertical Shift If g(x) is a function and k is a positive constant, then the graph of y = g(x) + k is the graph of y = g(x) shifted vertically upward by k units. y = g(x) − k is the graph of y = g(x) shifted vertically downward by k units. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

3. Vertical Shift: The Heating Schedule for an Office Building Example 1 To save money, an office building is kept warm only during business hours. At midnight (t = 0), the building’s temperature (H) is 50 ◦ F. This temperature is maintained until 4 am. Then the building begins to warm up so that by 8 am the temperature is 70 ◦ F. At 4 pm the building begins to cool. By 8 pm, the temperature is again 50 ◦ F. Suppose that the building’s superintendent decides to keep the building 5 ◦ F warmer than before. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Graph of the original heating schedule Graph of the new heating schedule obtained by shifting original graph upward by 5 units. t (hours after midnight) t H( ◦ F) H = f(t) H = p(t), new schedule H = f(t), original schedule

4. Vertical Shift: The Heating Schedule for an Office Building Example 2 What is the relationship between the formula for f(t), the original heating schedule and p(t), the new heating schedule? Solution The temperature under the new schedule, p(t), is always 5 ◦ F warmer than the temperature under the old schedule, f(t). Thus, The relationship between the formulas for p and f is given by the equation p(t) = f(t) + 5. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

5. Horizontal Shift If g(x) is a function and k is a positive constant, then the graph of y = g(x + k) is the graph of y = g(x) shifted horizontally to the left by k units. y = g(x − k) is the graph of y = g(x) shifted horizontally to the right by k units. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

6. Vertical Shift: The Heating Schedule for an Office Building Example 4 The superintendent then changes the original heating schedule to start two hours earlier. The building now begins to warm at 2 am instead of 4 am, reaches 70 ◦ F at 6 am instead of 8 am, begins cooling off at 2 pm instead of 4 pm, and returns to 50 ◦ F at 6 pm instead of 8 pm. How are these changes reflected in the graph of the heating schedule? Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Graph of the new heating schedule obtained by shifting original graph left by 2 units. t (hours after midnight) H( ◦ F) H = q(t), new schedule H = f(t), original schedule

7. Vertical Shift: The Heating Schedule for an Office Building Example 5 In Example 4 the heating schedule was changed to 2 hours earlier, shifting the graph horizontally 2 units to the left. Find a formula for q, this new schedule, in terms of f, the original schedule. Solution The old schedule always reaches a given temperature 2 hours after the new schedule. Thus, The relationship between the formulas for q and f is given by the equation q(t) = f(t + 2). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

8. Inside Versus Outside Changes Since the horizontal shift in the heating schedule, q(t) = f(t + 2), involves a change to the input value, it is called an inside change to f. Similarly the vertical shift, p(t) = f(t) + 5, is called an outside change because it involves changes to the output value. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

9. Combining Horizontal and Vertical Shifts Example 9 A graph of f(x) = x 2 is shown in blue. Define g by shifting the graph of f to the right 2 units and down 1 unit; the graph of g is shown in red. Find a formula for g in terms of f. Find a formula for g in terms of x. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y = f(x) y = g(x) The graph of g is the graph of f shifted to the right 2 units and down 1 unit, so a formula for g is g(x) = f(x − 2) − 1. Since f(x) = x 2, we have f(x − 2) = (x − 2) 2. Therefore, g(x) = (x − 2) 2 − 1.

10. 6.2 REFLECTIONS AND SYMMETRY Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

11. A Formula for a Reflection For a function f: The graph of y = −f(x) is a reflection of the graph of y = f(x) about the x-axis. The graph of y = f(−x) is a reflection of the graph of y = f(x) about the y-axis. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

12. Vertical Symmetry Example 1 (b) The graph shows a function f(x) in blue and its reflection through the x-axis in red (g(x)). The table shows values for f(x) and g(x). Write a formula for g(x) in terms of f(x) Solution (b) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally y = f(x) y = g(x) xf (x)g(x)g(x) -31 -22 4-4 08-8 116-16 232-32 364-64 x y When a point is reflected vertically about the x-axis, the x-value stays fixed, while the y- value changes sign. Algebraically, this means g(x) = − f (x).

13. Horizontal Symmetry Example 1 (b) The graph shows a function f(x) in blue and its reflection through the y-axis in red (h(x)). The table shows values for f(x) and h(x). Write a formula for h(x) in terms of f(x) Solution (b) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally xf (x)h(x)h(x) -3164 -2232 416 088 1 4 2322 3641 x y When a point is reflected horizontally about the y-axis, the y- value stays fixed, while the x-value changes sign. Algebraically, this means h(x) = f (− x). 64 32 y = h(x) y = f(x) 16

14. Horizontal and Vertical Reflection Combined Example 1 (c) The graph shows a function f(x) in blue and its reflection through both the x-axis the y-axis in red (k(x)). The table shows values for f(x) and k(x). Write a formula for k(x) in terms of f(x) Solution (c) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally xf (x)k(x)k(x) -31-64 -22-32 4-16 08-8 116-4 232-2 364 x y When a point is reflected about both the x-axis and the y- axis, both the x- value and the y- value change signs. Algebraically, this means k(x) = − f (− x). 64 32 y = k(x) y = f(x) -32 -64

15. Even Functions Are Symmetric About the y-Axis If f is a function, then f is called an even function if, for all values of x in the domain of f, f(−x) = f(x). The graph of f is symmetric about the y-axis. If f is a function, then f is called an even function if, for all values of x in the domain of f, f(−x) = f(x). The graph of f is symmetric about the y-axis. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

16. Example of an Even Function Consider the function f(x) = x 2 – 4. Note that f(– x) = (– x) 2 – 4 = x 2 – 4 So f(x) = f(–x). This means that f(x) is an even function, as is verified in the graph by its symmetry with respect to the y-axis Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Graph of y = f(x) = x 2 – 4

17. Odd Functions Are Symmetric About the Origin If f is a function, then f is called an odd function if, for all values of x in the domain of f, f(−x) = − f(x). The graph of f is symmetric about the origin. If f is a function, then f is called an odd function if, for all values of x in the domain of f, f(−x) = − f(x). The graph of f is symmetric about the origin. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

18. Example of an Odd Function Consider the function f(x) = x 3. Note that f(–x) = (– x) 3 = – x 3 So f(x) = – f(–x). This means that f(x) is an odd function, as is verified in the graph by its symmetry with respect to the origin. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Graph of y = f(x) = x 3

19. Combining Shifts and Reflections Consider the function f(x) = 1/x. Consider the reflection of this function through the x- axis together with a shift of 4 units in the positive vertical direction g(x) = 4 – 1/x Notice how both functions are asymptotic to the y-axis, but the horizontal asymptote for the transformed function has shifted up 4 units. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Graph of y = f(x) = 1/x and y = g(x) = 4 – 1/x Graph of y = f(x) = 1/x and y = g(x) = 4 – 1/x

20. 6.3 VERTICAL STRETCHES AND COMPRESSIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

21. Formula for Vertical Stretch or Compression If f is a function and k is a constant, then the graph of y = k · f(x) is the graph of y = f(x) Vertically stretched by a factor of k, if k > 1. Vertically compressed by a factor of k, if 0 < k < 1. Vertically stretched or compressed by a factor |k| and reflected across x-axis, if k < 0. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

22. Vertical Stretch: An Amplifier An amplifier takes a weak signal from a recording and transforms it into a stronger signal to power a set of speakers. Notice that the wave crests of the amplified signal are 3 times as high as those of the original signal; similarly, the amplified wave troughs are 3 times deeper than the original wave troughs. The amplifier has boosted the strength of the signal by a factor of 3. If f is the original signal function and V is the amplified signal function, then V(t) = 3 ・ f(t). As expected, a vertical stretch of the graph of f(t) corresponds to an outside change in the formula. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally The figure shows a graph of an audio signal (in volts) as a function of time, t, both before and after amplification. Amplified signal V(t) Original signal f(t) t, time Signal strength (volts)

23. Negative Stretch Factor The Figure gives a graph of a function y = f(x), together with a graph of y = − 2 · f(x) and the intermediate transformation y = 2 · f(x). The stretch factor is k = −2. We think of y = −2 f(x) as a combination of two separate transformations of y = f(x). First, the graph is stretched by a factor of 2, then it is reflected across the x-axis. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Final y = -2 f(x) Original y = f(x) Intermediate y = 2 f(x)

24. Stretch Factors and Average Rates of Change If g(x) = k · f(x), then on any interval, Average rate of change of g = k · (Average rate of change of f). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

25. Stretch Factors and Average Rates of Change Exercise 26 The function s(t) gives the distance (miles) in terms of time (hours). If the average rate of change of s(t) on 0 ≤ t ≤ 4 is 70 mph, what is the average rate of change of ½ s(t) on this interval? Solution The average rate of change of ½ s(t) on the interval 0 ≤ t ≤ 4 is ½ the average rate of change of s(t) on that interval, so it is ½ of 70 or 35 mph. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

26. Example 3 The function y = f(x) has the graph shown: Graph the function g(x) = − ½ f(x + 3) − 1. Solution To combine several transformations, always work from inside the parentheses outward. The graphs corresponding to each step are shown. Step 1: shift 3 units to the left Step 2: vertically compress by ½ Step 3: reflect about the x-axis Step 4: shift down 1 unit Combining Stretches and Shifts Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Original y = f(x) Step 1: y = f(x+3) Step 2: y = ½ f(x+3) Step 3: y =- ½ f(x+3) Step 4: y = -½ f(x+3) - 1

27. 6.4 HORIZONTAL STRETCHES AND COMPRESSIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

28. Formula for Horizontal Stretch or Compression If f is a function and k a positive constant, then the graph of y = f(k x) is the graph of f Horizontally compressed by a factor of 1/k if k > 1, Horizontally stretched by a factor of 1/k if k < 1. If k < 0, then the graph of y = f(kx) also involves a horizontal reflection about the y-axis. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

29. A Horizontal Stretch Example 1 The values and graph of the function f(x) are shown in blue. Make a table and a graph of the function g(x) = f(½ x). Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally xf(x)xg(x) -30-60 -22-42 0-20 00 1020 2 4 3161 f(x)f(x)g(x) = f(½ x)

30. Examples: Horizontal Stretch or Compression Example 3 Match the functions f(t) = e t, g(t) = e 0.5t, h(t) = e 0.8t, j(t) = e 2t with the graphs Solution Since the function j(t) = e 2t climbs fastest of the four and g(t) = e 0.5t climbs slowest, graph A must be j and graph D must be g. Similarly, graph B is f and graph C is h. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally ABCD

31. 6.5 COMBINING TRANSFORMATIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

32. Multiple Inside Changes Example 2 (a) Rewrite the function y = f(2x − 6) in the form y = f(B(x − h)). (b) Use the result to describe the graph of y = f(2x − 6) as the result of first applying a horizontal stretch or compression to the graph of f and then applying a horizontal shift. What is the stretch/compression factor? What is the shift? Solution (a) We have f(2x − 6) = f(2(x − 3)). (b) Thus, the graph of y = f(2x − 6) can be obtained from the graph of y = f(x) by first horizontally compressing it by a factor of ½ and then shifting horizontally right by h = 3 units. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

33. Graphical Demonstration of Example 2: f(2x – 6) Multiple Inside Changes Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Final y = f(2(x – 3))=f(2x – 6) Original y = f(x) Intermediate y = f(2x)

34. Multiple Outside Changes Example 2 (a) The graph of a function is obtained from the graph of y = f(x) by first stretching vertically by a factor of 3 and then shifting upward by 2. Give a formula for the function in terms of f. (b) The graph of a function is obtained from the graph of y = f(x) by first shifting upward by 2 and then stretching vertically by a factor of 3. Give a formula for the function in terms of f. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Original y = f(x) Intermediate y = 3 f(x) Intermediate y = f(x) + 2 Final y = 3 f(x ) + 2 Final y = 3(f(x ) + 2) = 3f(x) + 6 (c) 3f(x) + 2 ≠ 3f(x) + 6 (a) (b)

35. Ordering Horizontal and Vertical Transformations For nonzero constants A, B, h and k, the graph of the function y = A f(B (x − h)) + k is obtained by applying the transformations to the graph of f(x) in the following order: Horizontal stretch/compression by a factor of 1/|B| Horizontal shift by h units Vertical stretch/compression by a factor of |A| Vertical shift by k units If A < 0, follow the vertical stretch/compression by a reflection about the x-axis. If B < 0, follow the horizontal stretch/compression by a reflection about the y-axis. For nonzero constants A, B, h and k, the graph of the function y = A f(B (x − h)) + k is obtained by applying the transformations to the graph of f(x) in the following order: Horizontal stretch/compression by a factor of 1/|B| Horizontal shift by h units Vertical stretch/compression by a factor of |A| Vertical shift by k units If A < 0, follow the vertical stretch/compression by a reflection about the x-axis. If B < 0, follow the horizontal stretch/compression by a reflection about the y-axis. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

36. Multiple Transformations Example 5 (a) Let y = 5(x + 2) 2 + 7. Determine the values of A, B, h, and k when y is put in the form y = Af(B(x − h)) + k with f(x) = x 2. List the transformations applied to f(x) = x 2 to give y = 5(x + 2) 2 + 7. (b) Sketch a graph of y = 5(x + 2) 2 + 7, labeling the vertex. Solution Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally (a)If f(x) = x 2, we see y = 5(x + 2) 2 + 7 = 5f(x + 2) + 7, so A = 5, B = 1, h = −2, and k = 7. Based on the values of these constants, we carry out these steps: Horizontal shift 2 to the left of the graph of f(x) = x 2. Vertical stretch of the resulting graph by a factor of 5. Vertical shift of the resulting graph up by 7. Since B = 1, there is no horizontal compression, stretch, or reflection. y=x2y=x2 y = 5(x + 2) 2 + 7 (0,0) (-2,7) (b)