Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 103: 74

Answer

$a)$ $x$-intercepts: $\pm4;$ $y$-intercept: None $b)$ $x$-intercept: $4;$ $y$-intercept: $8$

Work Step by Step

$a)$ $y=\sqrt{x^{2}-16}$ To find the $x$-intercept, set $y$ equal to $0$ and solve for $x$: $y=\sqrt{x^{2}-16}$ $0=\sqrt{x^{2}-16}$ $(\sqrt{x^{2}-16})^{2}=0^{2}$ $x^{2}-16=0$ $x^{2}=16$ $x=\pm\sqrt{16}$ $x=\pm4$ To find the $y$-intercepts, set $x$ equal to $0$ and solve for $x$: $y=\sqrt{x^{2}-16}$ $y=\sqrt{0^{2}-16}$ $y=\pm\sqrt{-16}$ Since solving for $y$ yields a complex solution, this equation does not have $y$-intercept $b)$ $y=\sqrt{64-x^{3}}$ To find the $x$-intercept, set $y$ equal to $0$ and solve for $x$: $y=\sqrt{64-x^{3}}$ $0=\sqrt{64-x^{3}}$ $(\sqrt{64-x^{3}})^{2}=0^{2}$ $64-x^{3}=0$ $x^{3}=64$ $x=\sqrt[3]{64}$ $x=4$ To find the $y$-intercept, set $x$ equal to $0$ and solve for $y$: $y=\sqrt{64-x^{3}}$ $y=\sqrt{64-(0)^{3}}$ $y=\sqrt{64}$ $y=8$
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