How do you find vertical, horizontal and oblique asymptotes for #f(x) = (3x^2+2x-1)/( x^2-4)#?

Question

Sophie Adams · Accepted Answer

#"vertical asymptotes at "x=+-2#
#"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^2-4=0rArr(x-2)(x+2)=0# #rArrx=+-2" 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+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)# as #xto+-oo,f(x)to(3+0-0)/(1-0)# #rArry=3" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 2) hence there are no oblique asymptotes. graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the vertical asymptotes, factor the denominator and identify any values of $ x $ that make the denominator zero. These values represent vertical asymptotes.

To find horizontal asymptotes:
1. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $.
2. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.
3. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find oblique asymptotes (slant asymptotes), perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient represents the equation of the oblique asymptote. If the degree of the numerator is greater than the degree of the denominator by exactly one, there is an oblique asymptote.

For the given function $ f(x) = \frac{{3x^2+2x-1}}{{x^2-4}} $:
- Vertical asymptotes occur where the denominator equals zero, so $ x = 2 $ and $ x = -2 $.
- There's no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator.
- To find the oblique asymptote, perform polynomial division: $ (3x^2 + 2x - 1) \div (x^2 - 4) $. The quotient represents the equation of the oblique asymptote.