Exercise 41 Page 522

To find the horizontal asymptote, we first need to compare the degrees of the polynomials in the numerator and denominator. y=x^2-3x+2/x^2+6x+5 We see that both the numerator and denominator have a degree of 2. Therefore, the horizontal asymptote is the ratio of the coefficient of the term with a greatest degree in the numerator to the coefficient of the term with a greatest degree in the denominator. y=( 1)x^2-3x+2/( 1)x^2+6x+5 ⇒ y=1/1=1 Therefore, the horizontal asymptote is y=1. To find the vertical asymptote, we need to factor the numerator and denominator. x^2-3x+2/x^2+6x+5 ⇔ (x-2)(x-1)/(x+5)(x+1) We see that the numerator and denominator have no common zeros. This means that the function has two vertical asymptotes at the zeros of the denominator, x=- 5 and x=- 1. Horizontal Asymptote: & y=1 Vertical Asymptote: & x=- 5, x=- 1 The student thinks that the zeros of the numerator are the vertical asymptotes and that the zeros of the denominator are the horizontal asymptotes. However, we have shown that this is not the case.