Algebra and Trigonometry 10th Edition

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

Chapter 3 - 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 248: 14


Standard form: $f(x)=(x-4)^2-16$ Vertex: $(4,-16)$ Axis of symmetry: $x=4$. The x-intercepts: $(0,0)$ and $(8,0)$

Work Step by Step

$f(x)=x^2-8x~~$ ($a=1,~b=-8,~c=0$) $f(x)=x^2-8x+16-16$ $f(x)=x^2-2(4)x-4^2-16$ $f(x)=(x-4)^2-16$ $-\frac{b}{2a}=-\frac{-8}{2(1)}=4$ $f(3)=4^2-8(4)=16-32=-16$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(4,-16)$ So, the axis of symmetry is $x=-\frac{b}{2a}=4$. The x-intercepts: $f(x)=0$ $x^2-8x=0$ $x(x-8)=0$ $x=0$ or $x-8=0~→~x=8$ So, the x-intercept points are: $(0,0)$ and $(8,0)$
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