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: 56


$\left(-6,3 \right)$ and $\left(4,33\right)$

Work Step by Step

To find the points of intersection of $f(x)$ and $g(x)$, solve $f(x)=g(x)$: \begin{align*} x^2+5x-3&=2x^2+7x-27\\ 0&=-x^2-5x+3+2x^2+7x-27\\ 0&=x^2+2x-24 \end{align*} By Factoring: $$0=(x+6)(x-4)$$ Use the Zero-Product Property by equating each factor to zero,then solve each equation to obtain: \begin{align*} x+6 &=0 &\text{ or }& &x-4=0\\ x &=-6 &\text{ or }& &x =4\\ \end{align*} To find the y-coordinates of the points of intersection, evaluate either of the two functions at $x=-6$ and $x=4$ to obtain: $f(-6)=(-6)^2+5(-6)-3=3$ $f(4)=(4)^2+5(4)-3=33$ Thefore, the points of intersection are: $\left(-6,3 \right)$ and $\left(4,33\right)$
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