Stitz-Zeager_College_Algebra_e-book

A y s x 1 y 3 y 0 3 3 2 3 2 1 3 0 0

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Unformatted text preview: ˆ’2 can be thought of as a two step process: dividing by 2 which compresses the graph horizontally by a factor of 2 followed by dividing (multiplying) by βˆ’1 which causes a reflection across the y -axis. We summarize the results in the table below. 1.8 Transformations 101 (a, f (a)) (βˆ’4, βˆ’3) a βˆ’2x + 1 = a βˆ’4 βˆ’2x + 1 = βˆ’4 (βˆ’2, 0) βˆ’2 βˆ’2x + 1 = βˆ’2 (0, 3) 0 βˆ’2x + 1 = 0 (2, 0) 2 βˆ’2x + 1 = 2 (4, βˆ’3) 4 βˆ’2x + 1 = 4 x 5 2 x= 3 2 x= 1 2 1 x = βˆ’2 x = βˆ’3 2 x= 3 Next, we take each of the x values and substitute them into g (x) = βˆ’ 2 f (βˆ’2x + 1) + 2 to get the corresponding y -values. Substituting x = 5 , and using the fact that f (βˆ’4) = βˆ’3, we get 2 g 5 2 3 =βˆ’ f 2 βˆ’2 5 2 3 3 9 13 + 1 + 2 = βˆ’ f (βˆ’4) + 2 = βˆ’ (βˆ’3) + 2 = + 2 = 2 2 2 2 3 We see the output from f is first multiplied by βˆ’ 2 . Thinking of this as a two step process, multiplying by 3 then by βˆ’1, we see we have a vertical stretch by a factor of 3 followed by a 2 2 reflection across the x-axis. Adding 2 results in a vertical shift up 2 units. Continuing in this manner, we get the table below. x g (x) 5 2 3 2 1 2 βˆ’1 2 βˆ’3 2 13 2 2 βˆ’5 2 2 13 2 (x, g (x)) 5 13 2, 2 3 2, 2 1 5 2, βˆ’2 βˆ’1, 2 2 βˆ’ 3 , 13 22 To graph g , we plot each of the points in the table above and connect them in the same order and fashion as the points to which they correspond. Plotting...
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