What is the equation of the oblique asymptote #f(x)=(x^2-x-2)/(x+1)#? Precalculus Real Zeros of Polynomials Long Division of Polynomials 1 Answer George C. Jul 18, 2015 #f(x) = (x^2-x-2)/(x+1) = ((x-2)(x+1))/(x+1) = x-2# So #f(x)# is a straight line (with excluded point #(-1,-3)#). It has no asymptote. Explanation: #(x^2-x-2)/(x+1)# #= (x^2+x-2x-2)/(x+1)# #= (x(x+1)-2(x+1))/(x+1)# #= ((x-2)(x+1))/(x+1) = x-2# with exclusion #x != -1# Answer link Related questions What is long division of polynomials? How do I find a quotient using long division of polynomials? What are some examples of long division with polynomials? How do I divide polynomials by using long division? How do I use long division to simplify #(2x^3+4x^2-5)/(x+3)#? How do I use long division to simplify #(x^3-4x^2+2x+5)/(x-2)#? How do I use long division to simplify #(2x^3-4x+7x^2+7)/(x^2+2x-1)#? How do I use long division to simplify #(4x^3-2x^2-3)/(2x^2-1)#? How do I use long division to simplify #(3x^3+4x+11)/(x^2-3x+2)#? How do I use long division to simplify #(12x^3-11x^2+9x+18)/(4x+3)#? See all questions in Long Division of Polynomials Impact of this question 2214 views around the world You can reuse this answer Creative Commons License