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: 57

Answer

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

Work Step by Step

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