Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 2 - Linear and Quadratic Functions - Section 2.3 Quadratic Functions and Their Zeros - 2.3 Assess Your Understanding - Page 146: 48


Zeros: $-1 +\dfrac{2\sqrt{3}}{3},-1 -\dfrac{2\sqrt{3}}{3}$ $x$-intercepts: $-1 +\dfrac{2\sqrt{3}}{3},-1 -\dfrac{2\sqrt{3}}{3}$

Work Step by Step

To find the zeros of a function $f$, solve the equation $f(x)=0$ The zeros of the function are also the $x-$intercepts. Let $F(x)=0$: $$3x(x+2)-1=0$$ Expanding the equation: $$3x^2+6x-1=0$$ Comparing $3x^2+6x-1=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a = 3, b=6 , c =-1$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (6)^2-4 \times 3 \times -1 = 48$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{-6\pm \sqrt{48}}{2\times 3}$$ $$x=\dfrac{-6\pm 4\sqrt{3}}{6}$$ $\therefore x =-1 +\dfrac{2\sqrt{3}}{3}\hspace{20pt} \text{or} \hspace{20pt} x=-1 -\dfrac{2\sqrt{3}}{3}$
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