Find the horizontal asymptote of each function.f(x)=−5x(2x+3)2f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2}f(x)=(2x+3)2−5x

Horizontal Asymptote at y = 0 y=0 y = 0

Find the horizontal asymptote of each function. f ( x ) = − 5 x ( 2 x + 3 ) 2 f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2} f ( x ) = ( 2 x + 3 ) 2 − 5 x

In this problem, we're asked to find the horizontal asymptote of the function f of x is equal to negative 5x over 2x plus 3 squared. So we want to look at both the degree of the numerator and the denominator, and the degree of my numerator is simply 1, and the degree of my denominator is 2 because this x will end up getting squared. So that tells me that the degree of my numerator is less than the degree of my denominator, which which tells me that I'm going to have a horizontal asymptote at y equals 0. And we're done. I'll see you in the next video.

Find the horizontal asymptote of each function. f ( x ) = − 5 x ( 2 x + 3 ) 2 f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2} f ( x ) = ( 2 x + 3 ) 2 − 5 x

Horizontal Asymptote at y = 0 y=0 y = 0

Horizontal Asymptote at y = − 5 4 y=-\frac54 y = − 4 5

Horizontal Asymptote at y = − 5 2 y=-\frac52 y = − 2 5

5. Rational Functions

Asymptotes

Precalculus

5. Rational Functions

Asymptotes

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Multiple Choice

Find the horizontal asymptote of each function.
$f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2}$

Horizontal Asymptote at $y=0$

Horizontal Asymptote at $y=-\frac54$

Horizontal Asymptote at $y=-\frac52$

Verified step by step guidance

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.

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Struggling with Precalculus?

Find the horizontal asymptote of each function.f(x)=−5x(2x+3)2f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2}f(x)=(2x+3)2−5x​

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Find the horizontal asymptote of each function.
$f\left(x\right)=\frac{-5x}{\left(2x+3\right)^2}$