OKAY, people are telling my friend that the horizontal asymptote is the vertical asymptote?

Question

i'm asking for her because she already asked but got more confused instead.
HERE'S A PROBLEM: X-3/2X+1
WE KNOW THE VERTICAL ASYMPTOTE IS -1/2. HOW DO YOU FIND THE HORIZONTAL ASYMPTOTE? TELL THE ANSWER AND WHY PLEASE.

Answer 1

Horizontal asymtote ..
When degree of numerator and denominator polynomial is same then HORIZONTAL asymtote is coefficient of leading coefficient of numerator divide by denominator ..
like here degree of numerator and denominator is same 1 .
coefficient of leading term of numerator is 1
coefficient of leading term of denominator is 2

Horizontal asymtote .. y = 1/2

Degree of numerator is greater than denominator ..then Horizontal asy. doesnt exist ...then SLANT asy exist
for example ( X^2 + 2x + 1) / (x -5 )

Degree of numerator is less than denominator
then Horizontal asymtote is y = 0
for example ( x - 5) / ( x^2 + 2x +1 )

Hope this will help you ..:)

Answer 2

Note that the numerator (x-3) and the denominator (2x+1) are of the same degree. When this occurs in a rational equation look for a horizontal asymptote. In this example, the horizontal asymptote is y=1/2.

One way to find it is to solve y=(x-3)/(2x+1) for x
2xy +y = x-3
y+3 =x - 2xy
x = (y+3)/(1-2y)
So you can see y=1/2 cause the denominator to go to 0 and thus you have an asymptote at that point.

An easier way is to simply divide the coefficients of x in the numerator and denominator and get 1/2.

Answer 3

I assume you mean

y = (x - 3) / (2x + 1)

The horizontal asymptote is the value the function approaches as x approaches ±∞.

y = (x - 3) / (2x + 1) = (1 - 3/x) / (2 + 1/x)

The value of this function approaches

y = (1 - 0) / (2 + 0) = 1/2 as x→∞

And you are correct, the vertical asymptote is x = -1/2.

Answer 4

Divide (x-3) / (2x+1). You get 1/2 with remainder -7/2. So this is equal to

1/2 - (7/2) / (2X+1).

The last term reduces to zero as x → ± infinity. So the horizontal asymptote is y=1/2.