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: 69


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

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