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: 13


Standard form: $f(x)=(x-3)^2-9$ Vertex: $(3,-9)$ Axis of symmetry: $x=3$ The x-intercepts: $(0,0)$ and $(6,0)$

Work Step by Step

$f(x)=x^2-6x~~$ ($a=1,~b=-6,~c=0$) $f(x)=x^2-6x+9-9$ $f(x)=x^2-2(3)x-3^2-9$ $f(x)=(x-3)^2-9$ $-\frac{b}{2a}=-\frac{-6}{2(1)}=3$ $f(3)=3^2-6(3)=9-18=-9$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(3,-9)$ So, the axis of symmetry is $x=-\frac{b}{2a}=3$. The x-intercepts: $f(x)=0$ $x^2-6x=0$ $x(x-6)=0$ $x=0$ or $x-6=0~→~x=6$ So, the x-intercept points are: $(0,0)$ and $(6,0)$
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