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: 51


$(0,3)$ and $(-6,3)$

Work Step by Step

To find the points of intersection of $f(x)$ and $g(x)$, set $f(x)=g(x)$ and solve the equation: \begin{align*}x^2+6x+3&=3\\ x^2+6x&=0\\ x(x+6)&=0 \end{align*} Use the Zero-Product Property by equating each factor to zero, then solve each equation:: \begin{align*} x =0 \hspace{10pt} &\text{ or }& \hspace{10pt} x+6=0\\ x =0 \hspace{10pt} &\text{ or }& \hspace{10pt} x=-6 \end{align*} To find the y-coordinates of the points of intersection, evaluate either of the two functions at $x=0$ and $x=-6$ to obtain: $g(0)=3$ $g(-6)=3$ Therefore, the points of intersection are $(0,3)$ and $(-6,3)$.
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