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: 81


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

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 $P(x)=0$: $$6x^2-x-2=0$$ Comparing $6x^2-x-2=0$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a = 6, b=-1 , c =-2$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (-1)^2-4 \times 6 \times -2 = 49$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{1\pm \sqrt{49}}{2\times 6}$$ $$x=\dfrac{1\pm 7}{12}$$ Thus, the zeros, which are also the $x$-intercepts, are $\dfrac{2}{3}$ and $-\dfrac{1}{2}$.
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