How do you find the vertical, horizontal or slant asymptotes for #(6x^2+2x-1) /( x^2-1)#?

Question

Elias Ackerman · Accepted Answer

vertical asymptotes at #x=+-1#
horizontal asymptote at y = 6

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes. solve: #x^2-1=0rArrx=+-1# #rArrx=-1" and " x=1" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by the highest power of x that is #x^2# #f(x)=((6x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-1/x^2)=(6+2/x-1/x^2)/(1-1/x^2)# as #xto+-oo,f(x)to(6+0-0)/(1-0)# #rArry=6" is the asymptote"# Slant 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 slant asymptotes. graph{(6x^2+2x-1)/(x^2-1) [-20, 20, -10, 10]}

Greyson Brown · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. In this case, the denominator is $ x^2 - 1 $. Setting it equal to zero gives $ x^2 - 1 = 0 $. Solving for x, we get $ x = 1 $ and $ x = -1 $. So, the vertical asymptotes are $ x = 1 $ and $ x = -1 $.

To find the horizontal asymptotes, compare the degrees of the numerator and denominator. Since both have the same degree (2), divide the leading coefficients. For the given function, the horizontal asymptote is the ratio of the leading coefficients, which is $ 6/1 = 6 $.

To find the slant asymptote, divide the numerator by the denominator using long division or synthetic division. After performing the division, the quotient will represent the slant asymptote, if it exists.