## Precalculus (10th Edition)

Published by Pearson

# Chapter 3 - Linear and Quadratic Functions - 3.3 Quadratic Functions and Their Properties - 3.3 Assess Your Understanding - Page 145: 15

#### Answer

$F$

#### Work Step by Step

RECALL: (1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens: (i) upward when $a \gt0$ and has its vertex as its minimum; or (ii) downward when $a\lt0$ and has its vertex as its maximum. (2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$ Let's compare $f(x)=x^2-2x+1$ to $f(x)=ax^2+bx+c$. We can see that $a=1$, $b=-2$, $c=1$. $a\gt0$, hence the graph opens up, hence it's vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{-2}{2\cdot(1)}=1.$ Thus, the minimum value is $f(1)=(1)^2-2(1)+1=0.$ The vertex is at $(1,0)$ and the graph opens up. Hence, the graph of the given function is graph $F$.

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