College Algebra (10th Edition)

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

Chapter 2 - Review Exercises - Page 195: 8

Answer

There are no x-intercepts. The y-intercept is (0,1) The equation has symmetry only with respect to the y-axis.

Work Step by Step

To find the x-intercept(s), we set y to 0 and solve for x: $0=x^4+2x^2+1$ $0=(x^2+1)^2$ $0=x^2+1$ $x^2=-1$ $\sqrt{x^2}=\sqrt{-1}$ It is not possible to take a square root of a negative number. That means this equation has no x-intercepts. To find the y-intercept(s), we set x to y and solve for y: $y=0^4+2(0)^2+1$ $y=1$ To test for symmetry with respect to the x-axis, we substitute y for -y and check if it equals the original equation: $(-y)=x^4+2x^2+1$ $-y=x^4+2x^2+1$ nope To test for symmetry with respect to the y-axis, we substitute x for -x and check if it equals the original equation: $y=(-x)^4+2(-x)^2+1$ $y=x^4+2x^2+1\checkmark$ 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: $(-y)=(-x)^4+2(-x)^2+1$ $-y=x^4+2x^2+1$ nope
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