Stitz-Zeager_College_Algebra_e-book

Theorem 16 horizontal scalings suppose f is a

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Unformatted text preview: f (4) + 3 = −2 + 3 = 1. Hence, (4, 1) is the corresponding point on the graph of m. If we closely examine the arithmetic, we see that we first multiply f (4) by −1, which corresponds to the reflection across the x-axis, and then we add 3, which corresponds to the vertical shift. If we define an intermediate function m1 (x) = −f (x) to take care of the reflection, we get y y 3 3 (4, 2) 2 2 (1, 1) 1 1 (0, 0) 1 2 3 4 x (0, 0) −1 −2 −2 1 2 3 4 x −1 reflect across x-axis y = f (x) = √ −− − − − −→ −−−−−− x multiply each y -coordinate by −1 (1, −1) (4, −2) √ y = m1 (x) = −f (x) = − x To shift the graph of m1 up 3 units, we set m(x) = m1 (x) + 3. Since m1 (x) = −f (x), when √ we put it all together, we get m(x) = m1 (x) + 3 = −f (x) + 3 = − x + 3. We see from the graph that the range of m is (−, 3]. 6 If we had done the reflection first, then j1 (x) = f (−x). Following this by a shift left would give us j (x) = √ j1 (x + 3) = f...
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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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