## College Algebra (10th Edition)

Published by Pearson

# Chapter 4 - Section 4.3 - Quadratic Functions and Their Properties - 4.3 Assess Your Understanding - Page 300: 57

See below.

#### Work Step by Step

Let's compare $f(x)=2x^2+12x-3$ to $f(x)=ax^2+bx+c$. We can see that a=2, b=12, c=-3. a>0, hence the graph opens up, hence its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot 2}=-3.$ Hence the minimum value is $f(-3)=2(-3)^2+12(-3)-3=-21.$

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