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Multiple Choice
Find the horizontal asymptote of each function. f(x)=(2x+3)2−5x
A
Horizontal Asymptote at y=0
B
Horizontal Asymptote at y=−45
C
Horizontal Asymptote at y=−25
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Verified step by step guidance
1
Identify the degrees of the polynomial in the numerator and the denominator. The numerator is -5x, which is a polynomial of degree 1. The denominator is (2x+3)^2, which is a polynomial of degree 2.
Compare the degrees of the numerator and the denominator. Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is y = 0.
If the degree of the numerator had been equal to the degree of the denominator, the horizontal asymptote would be the ratio of the leading coefficients.
If the degree of the numerator had been greater than the degree of the denominator, there would be no horizontal asymptote, but rather an oblique asymptote.
Conclude that for the given function f(x) = -5x / (2x+3)^2, the horizontal asymptote is y = 0.