College Algebra (10th Edition)

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

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding - Page 186: 31

Answer

(a) The circle has a center (0.5,-1) and a radius of 0.5 units long. (b) See below. (c) There are no x-intercepts. There is only one y-intercept: (0,-1).

Work Step by Step

First, we need to make sure that the equation is in standard form $(x-h)^2+(y-k)^2=r^2$ In this case, we need to complete the square to achieve the standard form: $x^2+y^2-x+2y+1=0$ $x^2-x+y^2+2y=-1$ $x^2-x+(\frac{1}{2})^2+y^2+2y+(\frac{2}{2})^2=-1+(\frac{1}{2})^2+(\frac{2}{2})^2$ $(x-0.5)^2+(y+1)^2=-1+0.25+1$ $(x-0.5)^2+(y+1)^2=0.25$ $(x-0.5)^2+(y+1)^2=0.5^2$ Now we can find the center (h,k) and radius. The x-intercepts are all points of a graph when y=0: $(x-0.5)^2+(0+1)^2=0.25$ $(x-0.5)^2+1=0.25$ $(x-0.5)^2=-0.75$ $\sqrt{(x-0.5)^2}=\sqrt{-0.75}$ Taking the square root of a negative number is not possible; therefore, the graph doesn't intercept the x-axis at any point. The y-intercepts are all points of a graph when x=0: $(0-0.5)^2+(y+1)^2=0.25$ $0.25+(y+1)^2=0.25$ $(y+1)^2=0$ $\sqrt{(y+1)^2}=\sqrt{0}$ There is one y-intercept: $y+1=0\rightarrow y=-1$
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