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: 53


$(-1,-1)$ and $\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$

Work Step by Step

To find the points of intersection of $f(x)$ and $g(x)$, solve $f(x)=g(x)$: \begin{align*} -2x^2+1&=3x+2\\ 0&=2x^2-1+3x+2\\ 0&=2x^2+3x+1 \end{align*} By Factoring: $$0=(x+1)(2x+1)$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation: \begin{align*} x+1 &=0 &\text{ or }& &2x+1=0\\ x &=-1 &\text{ or }& &2x =-1\\ x &=-1 &\text{ or }& &x=-\dfrac{1}{2}\\ \end{align*} To find the y-coordinates of the points of intersection, evaluate either of the two functions at $x=-1$ and $x=-\dfrac{1}{2}$ to obtain: $g(-1)=3(-1)+2=-1$ $g(-\dfrac{1}{2})=3(-\dfrac{1}{2})+2=\dfrac{1}{2}$ Therefore, the points of intersection are $(-1,-1)$ and $\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$.
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