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 147: 95

Answer

The point of intersection is $(-8, 180)$.

Work Step by Step

We want to find when $f(x) = g(x)$, so we set the two expressions equal to one another and solve: $3(x^2 - 4) = 3x^2 + 2x + 4$ Use the distributive property: $3x^2 - 12 = 3x^2 + 2x + 4$ Move all terms to the left side of the equation: $-2x - 16 = 0$ Factor out common terms: $-2(x + 8) = 0$ Divide both sides by $-2$: $x + 8 = 0$ Subtract $8$ from each side of the equation: $x = -8$ We can now plug this value into either $f(x)$ or $g(x)$ to find the points of intersection of the two graphs. Let's use $f(x)$: $f(x) = 3[(-8)^2 - 4]$ Evaluate exponents first: $f(x) = 3(64 - 4)$ Simplify what is in parentheses: $f(x) = 3(60)$ Multiply to simplify: $f(x) = 180$ The point of intersection is $(-8, 180)$.
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