Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 248: 28

Answer

Standard form: $y=[x-(-\frac{1}{2})]^2-\frac{121}{4}$ Vertex: $(-\frac{1}{2},-\frac{121}{4})$ Axis of symmetry: $x=-\frac{1}{2}$ The x-intercepts are: $(5,0)$ and $(-6,0)$
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Work Step by Step

Standard form: $y=a(x-h)^2+k$, in which $(h,k)$ is the vertex. $y=-(x^2+x-30)$ $y=-[x^2+2(\frac{1}{2})x+(\frac{1}{2})^2-(\frac{1}{2})^2-30]$ $y=-[(x+\frac{1}{2})^2-\frac{121}{4}]$ $y=[x-(-\frac{1}{2})]^2-\frac{121}{4}$ Vertex: $(-\frac{1}{2},-\frac{121}{4})$ Axis of symmetry: $x=-\frac{1}{2}$ $y=0$ $[x-(-\frac{1}{2})]^2-\frac{121}{4}=0$ $[x-(-\frac{1}{2})]^2=\frac{121}{4}$ $x+\frac{1}{2}=±\frac{11}{2}$ $x=5$ or $x=-6$ The x-intercepts are: $(5,0)$ and $(-6,0)$
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