Answer
$(-3,11),(4,-10)$
Work Step by Step
$y=x^2-4x-10$
$y=-x^2-2x+14$
$-x^2-2x+14=x^2-4x-10$
$-2x^2+2x+24=0$
$-2(x^2-x-12)=0$
$(x+3)(x-4)=0$
$x=-3,4$
$y=(-3)^2-4(-3)-10$
$y=11$
$y=(4)^2-4(4)-10$
$y=-10$
$(-3,11),(4,-10)$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.4 - Page 559: 5
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