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: 45

Answer

x-intercept: None y-intercept: (0, 13/4) Vertical asymptote: x = 2 Domain: (-∞, 2) U (2, ∞) Horizontal asymptote: y=3 Range: (3, ∞) Graph is below.

Work Step by Step

$r(x)=\frac{3x^2-12x + 13}{x^2-4x +4}$ 3x^2 - 12x + 13 = 0 3(x^2 - 4x + 4) = -13 + 12 3(x-2)^2 = -1 No x-intercept y-intercept is the ratio of the constants, which is 13/4 Thus, the y-intercept is at (0, 13/4) Vertical asymptotes are when the denominator is equal to 0 $x^2 - 4x + 4 = 0$ x = 2 So, the domain is from (-∞, 2) U (2, ∞) Horizontal asymptote is the ratio of the constants of the leading term (with equal degree) 3/1 = 3 Thus, the horizontal asymptote is at y=3 So the range is from (3, ∞) (no negative values are possible)
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