Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.5 - Graphs of Quadratic Equations - Exercise Set: 8

Answer

$\text{$x-$intercepts: } \{ 1,-3 \} $

Work Step by Step

Setting $y=0$ and solving for $x$ using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), then the $x-$intercept/s of the given equation, $ y=-x^2-2x+3 ,$ is/are \begin{array}{l}\require{cancel} -x^2-2x+3=0 \\\\ x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(-1)(3)}}{2(-1)} \\\\ x=\dfrac{2\pm\sqrt{4+12}}{-2} \\\\ x=\dfrac{2\pm\sqrt{16}}{-2} \\\\ x=\dfrac{2\pm4}{-2} .\end{array} Hence, the $x-$intercepts are \begin{array}{l}\require{cancel} \text{$x-$intercepts: } \{ 1,-3 \} .\end{array}
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