College Algebra (10th Edition)

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

Chapter 2 - Review Exercises: 13

Answer

center: $(0. 1)$ radius $=2$ x-intercepts: $-\sqrt3$ and $\sqrt3$ y-intercepts: $-1$ and $3$
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Work Step by Step

RECALL: The standard form of a circle's equation is given as: $(x-h)^2+(y-k)^2=r^2$ where $r$ = radius and $(h, k)$ is the center. The given equation can be written as: $(x-0)^2+(y-1)^2=2^2$ Thus, the given circle has: center: $(0, 1)$ radius = $2$ To graph the circle, perform the following steps: (1) Plot the center $(0, 1)$. (2) With a radius of 2 units, plot the following points: 2 units above the center: $(0, 3)$ 2 units below the center: $(0, -1)$ 2 units to the left of the center: $(-2, 1)$ 2 units to the right of the center: $(2, 1)$ (3) Connect the four point in step (2) (not including the center) using a smooth curve to form a circle. (refer to the attached image in the answer part above for the graph) The circle has the following y-intercepts: $3$ and $-1$. To find the x-intercept, set $y=0$ then solve for $x$ to obtain: $x^2 + (y-1)^2=4 \\x^2 + (0-1)^2=4 \\x^2+1=4 \\x^2=4-1 \\x^2=3 \\x = \pm\sqrt{3}$ Thus, the x-intercepts are $\sqrt3$ and $-\sqrt3$.
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