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: 48

Answer

x-intercept: None y-intercept: (0, 3/2) Vertical Asymptote: x=-1 Domain: (-∞, -1) U (-1, ∞) Horizontal Asymptote: y=1/2 Range: (1/2, ∞) See graph below

Work Step by Step

$r(x)=\frac{x^2 + 2x +3}{2x^2 + 4x + 2}$ $x^2 + 2x +3= 0$ $x^2 + 2x + 1 = -3 + 1$ No x-intercept y-intercept is the ratio of the constants, which is 3/2 Thus, the y-intercept is at (0, 3/2) Vertical asymptotes are when the denominator is equal to 0 $2x^2 + 4x + 2 = 0$ $2 (x+1)^2 = 0$ x = -1 So, domain is from (-∞, -1) U (-1, ∞) Horizontal asymptote is the ratio of the constants of the leading term (with equal degree) 1/2 Thus, the horizontal asymptote is at y=1/2 So the range is from (1/2, ∞)
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