## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 4 Quadratic Functions and Factoring - 4.7 Complete the Square - 4.7 Exercises - Quiz for Lessons 4.5-4.7 - Page 291: 15

#### Answer

The vertex form of the function is $f(x)=(x-5)^{2}+(-8)$. The vertex is $(5,-8)$.

#### Work Step by Step

$f(x)=x^{2}-10x+17\qquad$ ...write in form of $x^{2}+bx=c$ (add $-17$ to each side). $f(x)-17=x^{2}-10x\qquad$ ...square half the coefficient of $x$. $(\displaystyle \frac{-10}{2})^{2}=(-5)^{2}=25\qquad$ ...complete the square by adding $25$ to each side of the expression $f(x)-17+25=x^{2}-10x+25\qquad$ ... write $x^{2}-10x+25$ as a binomial squared. $f(x)+8=(x-5)^{2}\qquad$ ...add $-8$ to each side of the expression $f(x)=(x-5)^{2}-8\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$. $f(x)=(x-5)^{2}+(-8)$ The vertex form of a quadratic function is $y=a(x-h)^{2}+k$ where $(h,k)$ is the vertex of the function's graph. Here, $h=5,\ k=-8$, so the vertex is $(5,-8)$

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