## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 3 - 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 248: 16

#### Answer

Standard form: $f(x)=(x+1)^2$ Vertex: $=(-1,0)$ Axis of symmetry: $x=-1$. The x-intercept: $(-1,0)$

#### Work Step by Step

$f(x)=x^2+2x+1~~$ ($a=1,~b=2,~c=1$) $f(x)=x^2+2(1)x+1^2$ $f(x)=(x+1)^2$ $-\frac{b}{2a}=-\frac{2}{2(1)}=-1$ $f(-1)=(-1+1)^2=0$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(-1,0)$ So, the axis of symmetry is $x=-\frac{b}{2a}=-1$. The x-intercepts: $f(x)=0$ $(x+1)^2=0$ $x+1=0$ $x=-1$ So, the x-intercept point is: $(-1,0)$

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