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: 66


Zeros: $-\dfrac{7}{2},-1$ $x$-intercepts: $-\dfrac{7}{2},-1$

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 $f(x)=0$: $$(2x+5)^2-(2x+5)-6=0$$ Let $u=2x+5$, the original equation becomes $$u^2-u-6=0$$ By factoring $$(u+2)(u-3) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} u+2 &= 0 &\text{ or }& &u=3\\ u &= -2 &\text{ or }& &u=3\\ \end{align*} To solve for $x$, we use $u=2x+5$ $$\because u = 2x+5$$ $$\therefore 2x = u-5$$ $$\therefore x = \dfrac{u-5}{2}$$ For $u=-2$ $$x=\dfrac{-2-5}{2} $$ $$x= -\dfrac{7}{2}$$ For $u=3$ $$x=\dfrac{3-5}{2}$$ $$x= -1$$ $\therefore x = -\dfrac{7}{2},-1$
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