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: 65


Zeros: $-6,-5$ $x$-intercepts: $-6,-5$

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 $G(x)=0$: $$(x+2)^2+7(x+2)+12=0$$ Let $u=x+2$, the original equation becomes $$u^2+7u+12=0$$ By factoring $$(u+4)(u+3) = 0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} u+4 &= 0 &\text{ or }& &u+3=0\\ u &= -4 &\text{ or }& &u=-3\\ \end{align*} To solve for $x$, we use $u=x+2$ $$\because u = x+2$$ $$\therefore x = u-2$$ For $u=-4$ $$x=-4-2 $$ $$x= -6$$ For $u=-3$ $$x=-3-2$$ $$x= -5$$ $\therefore x = -6,-5$
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