Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 3 - Linear and Quadratic Functions - 3.5 Inequalities Involving Quadratic Functions - 3.5 Assess Your Understanding - Page 160: 8



Work Step by Step

We are given the inequality: $x^2+3x-10>0$ $f(x)=x^2+3x-10$ Determine the intercepts: $y$-intercept: $f(0)=0^2+3(0)-10=-10$ $x$-intercepts: $x^2+3x-10=0$ $(x+5)(x-2)=0$ $x+5=0$ or $x-2=0$ $x=-5$ or $x=2$ The $y$-intercept is -10; the $x$-intercepts are -5 and 2. Determine the vertex: $x_V=-\dfrac{b}{2a}=-\dfrac{3}{2(1)}=-1.5$ $y_V=f(x_V)=(-1.5)^2+3(-1.5)-10=-12.25$ The vertex is $V(-1.5,-12.25)$ Graph the function. The graph is above the $x$-axis for: $x>2$ The solution set is: $\{x|x>2\}=(-\infty,-5)\cup(2,\infty)$
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