How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(3x^2+2) / (x^2 -1)#?

Question

Sadie Alexander · Accepted Answer

vertical asymptotes at x = ± 1
horizontal asymptote at y = 3

The denominator of f(x) cannot be zero as this would make f(x) 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^2=1rArrx=±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)=((3x^2)/x^2+2/x^2)/(x^2/x^2-1/x^2)=(3+2/x^2)/(1-1/x^2)# as #xto+-oo,f(x)to(3+0)/(1-0)# #rArry=3" 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{(3x^2+2)/(x^2-1) [-20, 20, -10, 10]}

Serenity Jones · Answer

undefined To find the vertical asymptotes, we set the denominator equal to zero and solve for x. So, for the function f(x) = (3x^2 + 2) / (x^2 - 1), the vertical asymptotes occur when x^2 - 1 = 0. Solving this equation, we get x = ±1.

To find the horizontal asymptote, we look at the degrees of the polynomials in the numerator and denominator. Since the degree of the numerator (2) is equal to the degree of the denominator (also 2), we divide the leading coefficients of both polynomials. Thus, the horizontal asymptote is y = 3.

To find the slant asymptote (if it exists), we divide the numerator by the denominator using long division or polynomial division. After performing the division, if the quotient is a polynomial of degree greater than 1, then there is no slant asymptote. If the quotient is a polynomial of degree 1 (a linear polynomial), then the equation of the slant asymptote is given by that linear polynomial.

For f(x) = (3x^2 + 2) / (x^2 - 1), performing the division, we get a quotient of 3 and a remainder of 5. Since the quotient is a constant (a polynomial of degree 0), there is no slant asymptote.