College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Review Exercises - Page 195: 7

Answer

The x-intercepts are (-4,0) and (4,0) The y-intercepts are (0,-2) and (0,2) The equation has symmetry with respect to the x-axis, y-axis and origin.

Work Step by Step

To find the x-intercept, we set y to 0 and solve for x: $x^2+4(0)^2=16$ $x^2=16$ $\sqrt(x^2)=\sqrt(16)$ $x=\pm4$ To find the y-intercept, we set x to y and solve for y: $0^2+4y^2=16$ $4y^2=16$ $y^2=4$ $\sqrt(y^2)=\sqrt(4)$ $y=\pm2$ To test for symmetry with respect to the x-axis, we substitute y for -y and check if it equals the original equation: $x^2+4(-y)^2=16$ $x^2+4y^2=16 \checkmark$ To test for symmetry with respect to the y-axis, we substitute x for -x and check if it equals the original equation: $(-x)^2+4y^2=16$ $x^2+4y^2=16\checkmark$ To test for symmetry with respect to the origin, we substitute x for -x, substitute y for -y and check if it equals the original equation: $(-x)^2+4(-y)^2=16$ $x^2+4y^2=16\checkmark$
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