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3 years ago

What is the horizontal asymptote of the rational function f(x) = 3x / (2x - 1)?

Mathematics

2 answers:

-BARSIC- [3]3 years ago

6 0

Answer:

Step-by-step explain

Find the horizontal asymptote for f(x)=(3x^2-1)/(2x-1) :

A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of y=a_n/b_n if the degree of the numerator is the same as the degree of the denominator (where a_n,b_n are the leading coefficients of the numerator and denominator respectively when both are in standard form.)

If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.

For the given function, there is no horizontal asymptote.

We can find the slant asymptote by using long division:

(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))

The slant asymptote is y=3/2x+3/4

butalik [34]3 years ago

6 0

Answer:

y = 3/2.

Step-by-step explanation:

This is the ratio of the coefficients of the terms in x of highest degree of the function:

y = 3/2.

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3 years ago

Express 2x^2+3x/(2x+3)(x-2) - 6/x^2-x-2 As a single fraction in its simplest form.

Dmitry [639]

Answer:

(x + 3)/(x + 1) .

Step-by-step explanation:

2x^2 + 3x/(2x + 3)(x - 2) - 6 /x^2 - x - 2

= 2x^2 + 3x/(2x + 3)(x - 2) - 6 / (x - 2)(x + 1)

The LCM is (2x + 3)(x - 2)(x + 1) so we have

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--------------------------------------- = ------------------------------------

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