Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter F - Foundations: A Prelude to Functions - Section F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry - F.2 Assess Your Understanding - Page 17: 62

Answer

$(\pm2,0), (0,-8)$, no symmetry

Work Step by Step

Step 1. For x-intercept(s), let $y=0$, we have $x^2(x+2)-4(x+2)=0\longrightarrow (x+2)(x^2-4)=0\longrightarrow (x+2)(x+2)(x-2)=0\longrightarrow (\pm2,0)$ Step 2. For y-intercept(s), let $x=0$, we have $y=-8\longrightarrow (0,-8)$, Step 3. To test for x-axis symmetry, replace $(x,y)$ with $(x,-y)$, we have $-y=(x)^3+2(x)^2-4(x)-8$ which is different from the original equation, thus it is not symmetric with respect to the x-axis, Step 4. To test for y-axis symmetry, replace $(x,y)$ with $(-x,y)$, we have $y=(-x)^3+2(-x)^2-4(-x)-8$ which is different from the original equation, thus it is not symmetric with respect to the y-axis, Step 5. To test for origin symmetry, replace $(x,y)$ with $(-x,-y)$, we have $-y=(-x)^3+2(-x)^2-4(-x)-8$ which is different from the original equation, thus it is not symmetric with respect to the origin.
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