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: 54


$\left(-\dfrac{2}{3},-\dfrac{17}{3}\right)$ and $\left(4,41\right)$

Work Step by Step

To find the points of intersection of $f(x)$ and $g(x)$, solve $f(x)=g(x)$: \begin{align*} 3x^2-7&=10x+1\\ 3x^2-10x-7-1&=0\\ 0&=3x^2-10x-8 \end{align*} By Factoring: $$0=(3x+2)(x-4)$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} 3x+2 &=0 &\text{ or }& &x -4=0\\ 3x&=-2 &\text{ or }& &x=4\\ x&=-\dfrac{2}{3} &\text{ or }& &x =4\\ \end{align*} To find the $y$-coordinates of the points of intersection, evaluate either of the two functions at $x=-\dfrac{2}{3}$ and $x=4$ to obtain: $g(-\dfrac{2}{3})=10(-\dfrac{2}{3})+1=-\dfrac{17}{3}$ $g(4)=10(4)+1=41$ Therefore, the points of intersection are: $\left(-\dfrac{2}{3},-\dfrac{17}{3}\right)$ and $\left(4,41\right)$
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