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: 76

Answer

x-intercepts: -3 and 3 y-intercepts: -2 and 2

Work Step by Step

RECALL: (i) The x-intercept is the value of x when y=0. (ii) The y-intercept is the value of y when x=0. Based on the given graph of the function, it appears that the function has: x-intercepts: -3 and 3 y-intercepts: -2 and 2 Check if these are the actual intercepts by substituting each one to the given equation: For x=-3: $\\\frac{(-3^2)}{9}+\frac{0^2}{4}=1 \\\frac{9}{9}+0=1 \\1=1$ For x=3: $\\\frac{3^2}{9}+\frac{0^2}{4}=1 \\\frac{9}{9}+0=1 \\1=1$ Thus, -3 and 3 are x-intercepts of the equation. For y=-2: $\\\frac{0^2}{9}+\frac{(-2)^2}{4}=1 \\0+\frac{4}{4}=1 \\1=1$ For y=2: $\\\frac{0^2}{9}+\frac{2^2}{4}=1 \\0+\frac{4}{4}=1 \\1=1$ Thus, -2 and 2 are y-intercepts of the equation.
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