ch06 - Chapter 6 Exponential Functions Review of...

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Review of Prerequisite Skills 3. d. 2 –1 3 + –1 2 –2 = = 1 1 2 2 = 6 + 4 3 = 9 4 5. b. (i) y 1 transforms to y 2 by a vertical shift upwards of four units. (ii) y 1 transforms to y 3 by a vertical shift downwards of three units. These are called translations. c. The graph of y = x 2 2, shifted vertically upwards four units, becomes the graph of y = x 2 + 2. d. When a positive or negative constant is added to a function, it results in a vertical shift of the graph of the function. For a positive constant, the shift is upwards that many units and for a negative constant, the shift is downward that many units. y x y y y 1 3 1 2 + 1 4 1 3 1 2 + 1 4 1 3 6. a. b. (i) The graph of y 1 is vertically compressed by one-half to form the graph of y 2 . (ii) The graph of y 1 is stretched vertically by two and it is reflected in the x- axis. c. The graph of y = 3 x 2 + 25 is the vertical stretch by three of the graph of y = x 2 and is shifted upwards 25 units. d. For the function y = c f ( x ) where c is a constant, the function is a transformation of y = f ( x ). If c < 0, the graph of the function is reflected in the x- axis. If 0 < c < 1, the graph of the function is compressed by a factor of c . If c > 1, the graph of the function is stretched by a factor of c . 7. a. b. (i) The graph of y 1 is shifted to the right five units to the graph of y 2 . (ii) The graph of y 1 is shifted to the left three units and reflected in the x -axis. y x x x x 5 y y y 1 3 y x x y y 3 y 1 Chapter 6: Exponential Functions 147 Chapter 6 • Exponential Functions
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i. (0.3) 3 ÷ (0.3) 5 = (0.3) –2 = 1 3 0 –2 = 1 3 0 2 = 10 9 0 k. (3 2 ) 3 ÷ 3 –2 = 3 6 ÷ 3 –2 = 3 6+2 = 3 8 = 6561 o. (6 3 ) 4 ÷ 12 6 = 1 6 2 12 6 = 6 6 6 12 2 6 = 6 2 6 6 = 3 6 = 729 2. g. 5 2 x x 3 y 2 y –4 2 = 5 2 x y 3 2 + + 2 4 = 5 2 x y 5 6 h. π 4 x x y 2 y 3 = 4 π y x 3 2 – 1 1 = π x 4 y –2 = 4 π y x 2 k. ( a 2 b –1 ) –3 = ( a 2 1 b –1 ) 3 or = a –6 b 3 = a 6 1 b –3 or = b a 3 6 = b a 3 6 148 Chapter 6: Exponential Functions c. The graph of y = ( x + 6) 2 7 is the lateral shift of six units to the left and seven units vertically downwards. d. When a positive or negative constant is added to the independent variable in a function, there is a lateral shift or translation. If the number is positive, it causes a shift to the left; if the number is negative it causes a shift to the right. In order to keep the same y- value, if the number is negative, the x- value must be increased to compensate, or decreased if the number is positive. Exercise 6.1 1. a. (7 3 ) 2 ÷ 7 4 = 7 6 ÷ 7 4 = 7 6–4 = 7 2 = 49 b. (0.4) 5 ÷ (0.4) 3 = (0.4) 2 = 1.6 c. ( 3) 5 ( 3) 3 = ( 3) 8 = (3 1 2 ) 8 = 3 4 = 81 d. 25 3 2 = 5 3 = 125 e. ( 8) 2 3 = 3 8 2 = ( 2) 2 = 4 f. ( 2) 3 ( 2) 3 = [( 2)( 2)] 3 = 4 3 = 64 g. 4 –2 8 –1 = 4 1 2 1 8 = 1 1 6 1 2 6 = 1 1 6
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l. ( ab ) 4 a b 2 2 2 = a 4 b 4 a b 4 4 = a 0 b 8 = b 8 3. b. a 1 4 b 1 3 –2 = a 1 2 b 2 3 = = 3 b a 2 d. = = x 1 – 1 3 y 1 6 1 1 2 = x 2 3 y 1 2 e.
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