How do you find the vertical, horizontal or slant asymptotes for #f(x)=(x-3 )/ (2x-1 )#?

Question

Ariana Alvarez · Accepted Answer

vertical asymptote # x = 1/2 #
horizontal asymptote # y = 1/2 # Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero. solve: 2x - 1 = 0 #rArr x = 1/2" is the asymptote " # Horizontal asymptotes occur as #lim_(xto+-oo) f(x) to 0 # divide all terms on numerator/denominator by x #(x/x - 3/x )/((2x)/x - 1/x) = (1 - 3/x)/(2 - 1/x) # as # x to+-oo , 3/x" and " 1/x to 0 # #rArr y = 1/2" is the asymptote " # Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes. Here is the graph of f(x). graph{(x-3)/(2x-1) [-10, 10, -5, 5]}

Lincoln Anderson · Answer

undefined To find the vertical, horizontal, or slant asymptotes for $ f(x) = \frac{x - 3}{2x - 1} $:

1. **Vertical Asymptotes**:
   Vertical asymptotes occur where the denominator of the rational function equals zero and the numerator does not. So, set the denominator $ 2x - 1 $ equal to zero and solve for $ x $.

$ 2x - 1 = 0 $  
   $ 2x = 1 $  
   $ x = \frac{1}{2} $

Therefore, the vertical asymptote is at $ x = \frac{1}{2} $.

2. **Horizontal Asymptotes**:
   To find horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote.

In this case, the degrees are equal (both 1). So, divide the leading coefficients.

$ \frac{1}{2} $

Therefore, the horizontal asymptote is at $ y = \frac{1}{2} $.

3. **Slant Asymptotes**:
   Slant asymptotes occur when the degree of the numerator is one greater than the degree of the denominator. To find the slant asymptote, perform polynomial long division.

$ (x - 3) $ divided by $ (2x - 1) $:

$ x - 3 $ = $ \frac{1}{2} $($ 2x - 1 $) + $ \frac{1}{2} $

So, the quotient is $ \frac{1}{2}x + \frac{1}{2} $.

Therefore, the slant asymptote is $ y = \frac{1}{2}x + \frac{1}{2} $.