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: 5


Standard form: $f(x)=(x-3)^2-8$ Vertex: $(3,-8)$ Axis of symmetry: $x=3$. x-intercepts: $(3-2\sqrt 2,0)$ and $(3+2\sqrt 2,0)$

Work Step by Step

$f(x)=x^2-6x+1~~$ ($a=1,~b=-6,~c=1$) $f(x)=x^2-6x+9-9+1$ $f(x)=[x^2-2(3)x+3^2]-8$ $f(x)=(x-3)^2-8$ $-\frac{b}{2a}=-\frac{-6}{2(1)}=3$ $f(3)=(3-3)^2-8=-8$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(3,-8)$ So, the axis of symmetry is $x=-\frac{b}{2a}=3$. The x-intercepts: $f(x)=0$ $(x-3)^2-8=0$ $(x-3)^2=8$ $x-3=±2\sqrt 2$ $x=3±2\sqrt 2$ So, the x-intercept points are: $(3-2\sqrt 2,0)$ and $(3+2\sqrt 2,0)$
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