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: 20

Answer

$(3,-4)$

Work Step by Step

Completing the square of the right side of the given equation, $ y=x^2-6x+5 ,$ results to \begin{array}{l}\require{cancel} y=(x^2-6x)+5 \\\\ y=\left( x^2-6x+\left( \dfrac{-6}{2} \right)^2 \right)+5-\left( \dfrac{-6}{2} \right)^2 \\\\ y=\left( x^2-6x+\left( -3 \right)^2 \right)+5-\left( -3 \right)^2 \\\\ y=\left( x^2-6x+9 \right)+5-9 \\\\ y=\left( x-3 \right)^2-4 .\end{array} The vertex of the parabola whose quadratic equation is in the form $y=a(x-h)^2+k$ is $(h,k)$. Hence, the vertex of the equation above is \begin{array}{l}\require{cancel} (3,-4) .\end{array}
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