## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 3 - Review Exercises - Page 300: 6

#### Answer

Standard form: $f(x)=(x+\frac{5}{2})^2-\frac{41}{4}$ Vertex: $(-\frac{5}{2},-\frac{41}{4})$ Axis of symmetry: $x=-\frac{5}{2}$. x-intercepts: $(\frac{-5-\sqrt {41}}{2},0)$ and $(\frac{-5+\sqrt {41}}{2},0)$

#### Work Step by Step

$f(x)=x^2+5x-4~~$ ($a=1,~b=5,~c=-4$) $f(x)=x^2+5x+\frac{25}{4}-\frac{25}{4}-4$ $f(x)=[x^2+2(\frac{5}{2})x+(\frac{5}{2})^2]-\frac{41}{4}$ $f(x)=(x+\frac{5}{2})^2-\frac{41}{4}$ $-\frac{b}{2a}=-\frac{5}{2(1)}=-\frac{5}{2}$ $f(-\frac{5}{2})=(-\frac{5}{2}+\frac{5}{2})^2-\frac{41}{4}=-\frac{41}{4}$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(-\frac{5}{2},-\frac{41}{4})$ So, the axis of symmetry is $x=-\frac{b}{2a}=-\frac{5}{2}$. The x-intercepts: $f(x)=0$ $(x+\frac{5}{2})^2-\frac{41}{4}=0$ $(x+\frac{5}{2})^2=\frac{41}{4}$ $x+\frac{5}{2}=±\frac{\sqrt {41}}{2}$ $x=-\frac{5}{2}±\frac{\sqrt {41}}{2}=\frac{-5±\sqrt {41}}{2}$ So, the x-intercept points are: $(\frac{-5-\sqrt {41}}{2},0)$ and $(\frac{-5+\sqrt {41}}{2},0)$

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