Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 2 - Functions and Their Graphs - 2.1 Functions - 2.1 Assess Your Understanding - Page 59: 118

Answer

Intercepts $x$ intercepts: $(-16,0)$ and $(-8,0)$ Symmetrical in $y$ axis

Work Step by Step

$\left(x+12\right)^{2}+\ y^{2}=16$ $x$-intercept: $\left(x+12\right)^{2}+\ y^{2}=16$ $\left(x+12\right)^{2}+\ (0)^{2}=16$ $x+12 = \pm\sqrt {16}$ $x=-12\pm4$ $x=-16, -8$ $y$-intercept: $\left(x+12\right)^{2}+\ y^{2}=16$ $\left(0+12\right)^{2}+\ y^{2}=16$ $144 +y^2=16$ $y=\pm\sqrt {16-144}$ $y=\pm\sqrt {-128}$ There is no $y$ intercept Therefore, the $x$ intercepts are $(-16,0)$ and $(-8,0)$ Test for symmetry Symmetry for $x$ axis: $\left(x+12\right)^{2}+\ y^{2}=16$ $x^2 +24x +144 + y ^2 = 16$ $\left(-x+12\right)^{2}+\ y^{2}=16$ $x^2-24x+144+y^2=16$ Not symmetric about $x$ axis Symmetry for $y$ axis: $\left(x+12\right)^{2}+\ y^{2}=16$ $x^2 +24x +144 + y ^2 = 16$ $\left(x+12\right)^{2}+\ y^{2}=16$ $\left(x+12\right)^{2}+\ (-y)^{2}=16$ $\left(x+12\right)^{2}+\ y^{2}=16$ Symmetric about $y$ axis Symmetry for origin: $\left(x+12\right)^{2}+\ y^{2}=16$ $x^2 +24x +144 + y ^2 = 16$ $\left(x+12\right)^{2}+\ y^{2}=16$ $\left(-x+12\right)^{2}+\ (-y)^{2}=16$ $x^2 -24x +144 + y ^2 = 16$ Not symmetric in the origin Therefore, it is symmetrical in $y$ axis.
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