Stitz-Zeager_College_Algebra_e-book

# This line is called the least squares regression line

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Unformatted text preview: mple 2.4.4. Solve the following inequalities analytically using sign diagrams. Verify your answer graphically. 1. 2x2 ≤ 3 − x 3. 9x2 + 4 ≤ 12x 2. x2 > 2x + 1 4. 2x − x2 ≥ |x − 1| − 1 Solution. 1. To solve 2x2 ≤ 3 − x, we ﬁrst get 0 on one side of the inequality which yields 2x2 + x − 3 ≤ 0. We ﬁnd the zeros of f (x) = 2x2 + x − 3 by solving 2x2 + x − 3 = 0 for x. Factoring gives (2x + 3)(x − 1) = 0, so x = − 3 ,or x = 1. We place these values on the number line with 0 2 above them and choose test values in the intervals −∞, − 3 , − 3 , 1 , and (1, ∞). For the 2 2 interval −∞, − 3 , we choose2 x = −2; for − 3 , 1 , we pick x = 0; and for (1, ∞), x = 2. 2 2 Evaluating the function at the three test values gives us f (−2) = 3 > 0 (so we place (+) 3 above −∞, − 3 ; f (0) = −3 < 0 (so (−) goes above the interval − 2 , 1 ); and, f (2) = 7 2 (which means (+) is placed above (1, ∞)). Since we are solving 2x2 + x − 3 ≤ 0, we look for solutions to 2x2 + x − 3 < 0 as well as solutions for 2x2 + x − 3...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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