Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 9 - Quadratic Functions and Equations - Chapter Review - Page 604: 11


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Work Step by Step

We are given: $y=\frac{1}{2}x^2+2x-3$ Let's list some values: $x=-2 \rightarrow y=1$ $x=-1 \rightarrow y=\frac{-1}{2}$ $x=0 \rightarrow y=-3$ $x=1 \rightarrow y=\frac{-1}{2}$ $x=2 \rightarrow y=1$ The x-coordinate of the vertex is given by $x=\frac{-b}{2a}=\frac{-2}{2.(\frac{1}{2})}=-2$ Find the y-coordinate of the vertex $y=\frac{1}{2}(-2)^2+2(-2)-3$ Hence, the vertex is $(-2,-5)$.
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