Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.6 - Rational Expressions - 3.6 Exercises - Page 309: 54

Answer

x-intercept: (-6,0) (1,0) y-intercept: (0, 2) Vertical Asymptote: x=-3, 2 Domain: (-∞, -3) U (-3, 2) U (2, ∞) Horizontal Asymptote: y=2 Range: (-∞, ∞) See graph below

Work Step by Step

$r(x)=\frac{2x^2 + 10x - 12}{(x^2 + x - 6)}$ $2x^2 + 10x - 12 = 0$ $2 (x^2 + 5x - 6) = 0$ $2 (x+6)(x-1) = 0$ x = -6, 1 x-intercept: (-6,0) (1,0) y-intercept is the ratio of the constants, which is -12 / -6 = 2 Thus, the y-intercept is at (0, 2) Vertical asymptotes are when the denominator is equal to 0 $x^2 + x - 6 = 0$ $(x+3)(x-2) = 0$ x = -3, 2 So, domain is from (-∞, -3) U (-3, 2) U (2, ∞) Horizontal asymptote is the ratio of the constants of the leading term (with equal degree) 2/1 = 2 Thus, the horizontal asymptote is at y=2 So the range is from (-∞, ∞)
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