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=0y=0y=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)=−5x(2x+3)2f\left(x\right)=\frac{-5x}{\left(2x+3\$$right)^2$$}f(x)=(2x+3)2−5x

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

Horizontal Asymptote at y=−54y=-\frac54y=−45

Horizontal Asymptote at y=−52y=-\frac52y=−25

5. Rational Functions

Asymptotes

College Algebra

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 $y=-$\frac$54$

Horizontal Asymptote at $y=-$\frac$52$

Verified step by step guidance

Identify the degrees of the polynomial in the numerator and the denominator. The degree of the numerator is 1 (since the highest power of x is x^1), and the degree of the denominator is 2 (since the highest power of x is (2x)^2).

Recall the rule for finding horizontal asymptotes: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.

Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is y = 0.

Verify by considering the behavior of the function as x approaches infinity. As x becomes very large, the term with the highest degree in the denominator will dominate, causing the function to approach zero.

Conclude that the horizontal asymptote of the function f(x) = $\frac{-5x}{(2x+3)^2}$ is y = 0.

6:24m

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Struggling with College Algebra?

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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Asymptotes practice set

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