How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=((3x-2)(x+5))/((2x-1)(x+6))#?

Question

Christian Antunez · Accepted Answer

vertical asymptotes x = -6 , x#=1/2#
horizontal asymptote #y=3/2# For this rational function the denominator cannot be zero. This would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the values that x cannot be and if the numerator is also non-zero for these values of x then they must be vertical asymptotes. solve : (2x-1)(x +6 ) =0 #rArrx=-6,x=1/2# #rArrx=-6" and " x=1/2" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# Now #f(x)=((3x-2)(x+5))/((2x-1)(x+6))=(3x^2+13x-10)/(2x^2+11x-6)# divide terms on numerator/denominator by the highest exponent of x , that is #x^2# #((3x^2)/x^2+(13x)/x^2-10/x^2)/((2x^2)/x^2+(11x)/x^2-6/x^2)=(3+13/x-10/x^2)/(2+11/x-6/x^2)# as #xto+-oo,f(x)to(3+0-0)/(2+0-0)# #rArry=3/2" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no oblique asymptotes. graph{((3x-2)(x+5))/((2x-1)(x+6) [-10, 10, -5, 5]}

Caleb Alexander · Answer

undefined To find the vertical, horizontal, and oblique asymptotes of the function $ f(x) = \frac{(3x - 2)(x + 5)}{(2x - 1)(x + 6)} $, follow these steps:

1. Determine the vertical asymptotes by finding the values of $ x $ where the denominator of the function equals zero.
2. Find the horizontal asymptote by examining the behavior of the function as $ x $ approaches positive or negative infinity.
3. Determine the oblique asymptote by performing polynomial long division or synthetic division if the degree of the numerator is greater than or equal to the degree of the denominator.

Let's proceed with the steps:

1. Vertical Asymptotes:
   Set the denominators equal to zero and solve for $ x $:
   $ 2x - 1 = 0 $ and $ x + 6 = 0 $
   Solve for $ x $:
   $ 2x = 1 $ $ \Rightarrow $ $ x = \frac{1}{2} $
   $ x = -6 $
   Therefore, the vertical asymptotes are $ x = \frac{1}{2} $ and $ x = -6 $.

2. Horizontal Asymptote:
   To find the horizontal asymptote, examine the behavior of the function as $ x $ approaches positive or negative infinity. 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, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.
   In this case, the degrees of the numerator and the denominator are equal (both are 2). Therefore, the horizontal asymptote is the ratio of the leading coefficients:
   $ y = \frac{3}{2} $.

3. Oblique Asymptote:
   Perform polynomial long division or synthetic division to find the equation of the oblique asymptote if the degree of the numerator is greater than or equal to the degree of the denominator. In this case, the degree of the numerator (2) is less than the degree of the denominator (3), so there is no oblique asymptote.

Therefore, the vertical asymptotes are $ x = \frac{1}{2} $ and $ x = -6 $, the horizontal asymptote is $ y = \frac{3}{2} $, and there is no oblique asymptote.