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: 42


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

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 $g(x)=0$: $$2x^2+5x+3=0$$ Comparing $2x^2+5x+3$ to $ax^2+bx+c=0$ to find $a,b \text{ and } c$ $$\therefore a = 2, b=5 , c =3$$ Evaluating the discriminant $b^2-4ac$ $$b^2-4ac = (5)^2-4 \times 2 \times 3 = 1$$ The quadratic formula is given by: $$x= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$$ $$x= \dfrac{-5 \pm \sqrt{1}}{2\times 2}$$ $$x=\dfrac{-5 \pm 1}{4}$$ $\therefore x =-1\hspace{20pt} \text{or} \hspace{20pt} x=-\dfrac{3}{2}$
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