Algebra and Trigonometry 10th Edition

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

Chapter 3 - Review Exercises - Page 300: 7


Standard form: $f(x)=(x+4)^2-6$ Vertex: $(-4,-6)$ Axis of symmetry: $x=-4$. x-intercepts: $(4+\sqrt 6,0)$ and $(4-\sqrt 6,0)$

Work Step by Step

$f(x)=x^2+8x+10~~$ ($a=1,~b=8,~c=10$) $f(x)=x^2+8x+16-16+10$ $f(x)=[x^2+2(4)x+4^2]-6$ $f(x)=(x+4)^2-6$ $-\frac{b}{2a}=-\frac{8}{2(1)}=-4$ $f(-4)=(4-4)^2-6=-6$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(-4,-6)$ So, the axis of symmetry is $x=-\frac{b}{2a}=-4$. The x-intercepts: $f(x)=0$ $(x+4)^2-6=0$ $(x+4)^2=6$ $x+4=±\sqrt 6$ $x=4±\sqrt 6$ So, the x-intercept points are: $(4+\sqrt 6,0)$ and $(4-\sqrt 6,0)$
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