Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.6 - Further Graphing of Quadratic Functions - Exercise Set - Page 527: 79

Answer

Vertex: $(-5,-10)$ Opens upward x-intercepts: -1.8, -8.2 y-intercept: 15

Work Step by Step

$f(x)=x^2+10x+15$ $y=x^2+10x+15$ $y=x^2+10x+(10/2)^2-(10/2)^2+15$ $y=x^2+10x+5^2-5^2+15$ $y=(x+5)^2-25+15$ $y=(x+5)^2-10$ Vertex: $(-5,10)$ Coefficient of $x^2$ is positive, so graph opens up $y=0$ $y=(x+5)^2-10$ $0=(x+5)^2-10$ $0+10=(x+5)^2-10+10$ $10=(x+5)^2$ $\sqrt {10}=\sqrt {(x+5)^2}$ $±\sqrt 10 = x+5$ $±3.16 = x+5$ $3.16=x+5$ $3.16-5=x+5-5$ $-1.84 = x$ $x=-1.8$ $-3.16=x+5$ $-3.16-5=x+5-5$ $-8.16=x$ $x=-8.2$ $x=0$ $y=(x+5)^2-10$ $y=(0+5)^2-10$ $y=5^2-10$ $y=25-10$ $y=15$
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