Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.3 - Trigonometric Graphs - 5.3 Exercises - Page 429: 20

Answer

amplitude=1 period =$\pi$ graph (black): .

Work Step by Step

See p. 424 For $y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$, $(k>0)$ Amplitude: $|a|$, Period: $\displaystyle \frac{2\pi}{k}$, Horizontal shift: $b$ ------------------ $y=-\sin 2x$ $a=-1 , k=2, b=0$ So, amplitude=$|-1|=1$ period =$\displaystyle \frac{2\pi}{2}=\pi$ To graph, begin with $f(x)=\cos x,$ (red, dashed) horizontally compress by factor 2, (green, dashed) then reflect across the x-axis (black, solid line).
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