Stitz-Zeager_College_Algebra_e-book

# If we attempt to build a table directly we quickly

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Unformatted text preview: and wish to ﬁnd the point on y = j (x) = f (−x + 3) which corresponds to (4, 2). We set −x + 3 = 4 and solve. Our ﬁrst step is to subtract 3 from both sides to get −x = 1. Subtracting 3 from the x-coordinate 4 is shifting the point (4, 2) to the left. From −x = 1, we then multiply5 both sides by −1 to get x = −1. Multiplying the x-coordinate by −1 corresponds to reﬂecting the point about the y -axis. Hence, we perform the horizontal shift ﬁrst, then follow it with the reﬂection about the y -axis. Starting with √ f (x) = x, we let j1 (x) be the intermediate function which shifts the graph of f 3 units to the left, j1 (x) = f (x + 3). y y (1, 2) (4, 2) 2 2 (−2, 1) (1, 1) 1 1 (−3, 0) (0, 0) −4 −3 −2 −1 1 y = f (x) = 2 √ 3 4 x shift left 3 units −4 −3 −2 −1 1 2 −− − − − −→ −−−−−− x subtract 3 from each x-coordinate y = j1 (x) = f (x + 3) = 3 √ 4 x x+3 To obtain the function j , we reﬂect the graph of j1 about y -axis. Theorem 1.4 tells us we...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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