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

Question

Liam Brennan · Accepted Answer

vertical asymptotes x = -1 , x = 4
horizontal asymptote y = 0

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation equate the denominator to zero. solve: #x^2-3x-4 = 0 → (x-4)(x+1) = 0# #rArr x = -1 and x = 4 " are asymptotes " # Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0# If the degree of the numerator is less than the degree of the denominator, as in this case, degree of numerator is 1 and degree of denominator is 2 then the equation of asymptote is y = 0 Here is the graph of the function. graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}

Chloe Alexander · Answer

undefined To find the asymptotes of $ f(x) = \frac{2x + 4}{x^2 - 3x - 4} $:

1. Check for vertical asymptotes by finding the values of $ x $ that make the denominator zero. These are the values that are not in the domain of the function.
   
2. Check for horizontal asymptotes by analyzing the behavior of the function as $ x $ approaches positive or negative infinity.

3. Determine if there are any slant (oblique) asymptotes by performing polynomial long division if the degree of the numerator is greater than or equal to the degree of the denominator.

4. Once you have found the vertical, horizontal, and slant asymptotes, state them as the equations of lines.