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: 8

Answer

Standard form: $f(x)=(x-4)^2-4$ Vertex: $(4,-4)$ Axis of symmetry: $x=4$. The x-intercepts: $(6,0)$ and $(2,0)$
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Work Step by Step

$f(x)=x^2-8x+12~~$ ($a=1,~b=-8,~c=12$) $f(x)=x^2-8x+16-16+12$ $f(x)=(x^2-2(4)x+4^2)-4$ $f(x)=(x-4)^2-4$ $-\frac{b}{2a}=-\frac{-8}{2(1)}=4$ $f(-1)=(4-4)^2-4=-4$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(4,-4)$ So, the axis of symmetry is $x=-\frac{b}{2a}=4$. The x-intercepts: $f(x)=0$ $(x-4)^2-4=0$ $(x-4)^2=4$ $x-4=±2$ $x=6$ or $x=2$ So, the x-intercept points are: $(6,0)$ and $(2,0)$
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