How do you find the asymptotes for #y = (3-4x )/( 6x + sqrt(4x^2-3x-5))#?

Question

Sadie Alexander · Accepted Answer

Horizontal: #y=-2/3#. The graph is real for #(-sqrt89+3)/8<=x<=(sqrt89+3)/8#.

To make y real, #(-sqrt89+3)/8<=x<=(sqrt89+3)/8#. #y =(-4+3/x)/(6+sqrt(4-3/x-1/x^2)) to -2/3#, as #x to +-oo#. Look at the dead ends #x = (3+-sqrt 89)/8= 1.554 and -0.804#, nearly. graph{y(6x+sqrt(4x^2-3x-5))+4x-3=0 [-5, 5, -2.5, 2.5]}

Elijah Allen · Answer

undefined To find the asymptotes of the function $ y = \frac{3 - 4x}{6x + \sqrt{4x^2 - 3x - 5}} $, we need to consider the behavior of the function as $ x $ approaches certain values.

1. Vertical Asymptotes:
   Vertical asymptotes occur where the denominator of the rational function becomes zero but the numerator does not. In this case, the denominator $ 6x + \sqrt{4x^2 - 3x - 5} $ becomes zero when $ x $ satisfies the equation $ 6x + \sqrt{4x^2 - 3x - 5} = 0 $. Solve this equation to find the values of $ x $ that create vertical asymptotes.

2. Horizontal Asymptotes:
   Horizontal asymptotes occur when $ x $ approaches positive or negative infinity. To find horizontal asymptotes, divide the highest degree term in the numerator by the highest degree term in the denominator. If the result is a constant, that constant is the horizontal asymptote.

3. Oblique Asymptotes (if applicable):
   If the degree of the numerator is exactly one greater than the degree of the denominator, there may be an oblique asymptote. To find this asymptote, perform long division of the numerator by the denominator and the quotient will be the equation of the oblique asymptote.

After determining these asymptotes, you can plot the function to visualize its behavior near these asymptotes.