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: 61

Answer

X-intercept(s): $ (\pm3,0),(-1,0)$, Y-intercept(s): $(0,-9)$, no symmetry

Work Step by Step

Step 1. For x-intercept(s), let $y=0$, we have $x^2(x+1)-9(x+1)\longrightarrow (x+1)(x^2-9)\longrightarrow (x+1)(x+3)(x-3)\longrightarrow (\pm3,0),(-1,0)$ Step 2. For y-intercept(s), let $x=0$, we have $y=-9\longrightarrow (0,-9)$, Step 3. To test for x-axis symmetry, replace $(x,y)$ with $(x,-y)$, we have $-y=(x)^3+(x)^2-9(x)-9$ 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 $y=(-x)^3+(-x)^2-9(-x)-9$ 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 $-y=(-x)^3+(-x)^2-9(-x)-9$ which is different from the original equation, thus it is not symmetric with respect to the origin.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.