Name: SOLUTIONS Date: Period: Assignment2A:CombiningTransformations Lippman/Rasmussen Answer the following problems from your Lippman/Ras mussen textbook with as much details, explanations, and work that is appropriate. 1.5: 16, 2626-28, 56, 99. 16. Tables of values for f ( x ) , g ( x ) , and h( x) are given below. Write g ( x ) and h( x) as transformations of f ( x ) . x -2 -1 0 1 2 f(x) -1 -3 4 2 1 x -3 -2 -1 0 1 g(x) -1 -3 4 2 1 3(4) = 6(4 + 8), x -2 -1 0 1 2 h(x) -2 -4 3 1 0 9(4) = 6(4) − 8 For the following, explain how the graph is a transformation of a toolkit function, then write an equation for each function graphed. 26. >(? ) = (? − 1) − 3 @ 27. >(? ) = √? + 3 − 1 28. >(? ) = |? + 2| + 2 56. f ( x) = x 2 horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. @ @ 8 1 4 >(? ) = C (? + 4)E − 3 = C ? + E − 3 D 3 3 99. Suppose you have a function y = f ( x ) such that the domain of f ( x ) is 1 ≤ x ≤ 6 and the range of f ( x ) is −3 ≤ y ≤ 5. [UW] a. What is the domain of f (2( x − 3)) ? Solutions on next page. b. What is the range of f ( 2( x − 3)) ? c. What is the domain of 2 f ( x) − 3 ? d. What is the range of 2 f ( x) − 3 ? e. Can you find constants B and C so that the domain of f ( B ( x − C )) is 8 ≤ x ≤ 9? f. Can you find constants A and D so that the range of Af ( x ) + D is 0 ≤ y ≤ 1? (a) This transformation will shift the graph horizontally to the right 3 units, and then compress the graph horizontally by a factor of L @ . To find the domain boundaries we need 1 = 2(? − 3) → ? = 3.5 6 = 2(? − 3) → ? = 6 So, the new domain is D. N ≤ 4 ≤ O. (b) The transformations applied to this function are horizontal shifts and stretches, which don’t affect the range, since the range is concerned with the y-values. So the range is unchanged. (c) The transformations applied to this function are vertical shifts and stretches, which don’t affect the domain, since the domain is concerned with the x-values. So the domain is unchanged. (d) This transformation will stretch the graph vertically by a factor of 2, and then shift the graph vertically down 3 units. The range will experience the same changes. So, the stretch changes the range to −6 ≤ Q ≤ 10, and then the shift changes the range to −R ≤ S ≤ T.. (e) We want to transform the domain 1 ≤ ? ≤ 6 into the domain 8 ≤ ? ≤ 9 using a horizontal shift and a horizontal stretch. So, we need to find the numbers B and C that give us 8 when we subtract C from 1 and multiply by B, and that also give us 9 when we subtract C from 6 and multiply by B . This can be represented as a system of equations: W(6 − X) = 9, 8 and W(1 − X) = 8. Solving this system gives Y = N , Z[\ ] = −DR. We can check that these L constants are correct by applying these changes > C^ (? + 39)E to the original function with the domain 1 ≤ ? ≤ 6, and we get a new domain of 8 ≤ ? ≤ 9, so it is correct. (f) The original range of −3 ≤ Q ≤ 5 a span of 8, and the new range of 0 ≤ Q ≤ 1 has a span L of 1. So the graph has to compress into _ of the space. After compressing it, the range ` ^ becomes − _ ≤ Q ≤ _. Now it is the right size, but not in the right location. To get it between 0 ` and 1, we need to move it up, shifting it vertically a distance of _, which is the same as adding ` _ a _ to the minimum and maximum. Now the range is _ ≤ Q ≤ _ which equals 0 ≤ Q ≤ 1. So, we L ` 8 D applied a compression of is _ and a shift of _, so o b = c and d = c.