How do you find the vertical, horizontal and slant asymptotes of: # f(x) =(x^2+6x-9)/(x-2)#?

Question

Caleb Alexander · Accepted Answer

V.A.: x=2, S.A.: y=x+8

The vertical asymptote depends on the domain of the function, then since it must be: #x!=2#, the vertical asymptote is the line #x=2# graph{(x^2+6x-9)/(x-2) [-10, 10, -5, 5]} In fact #lim_(x rarr2)(x^2+6x-9)/(x-2)=7/0=oo# There are no horizontal asymptote, because #lim_(x rarroo)(x^2+6x-9)/(x-2)=oo# Maybe the function has a slant asymptote, so let's calculate #m=lim_(x rarroo)(x^2+6x-9)/(x-2)*1/x=1# and #n=lim_(x rarr2)(x^2+6x-9)/(x-2)-mx=# #=lim_(x rarr2)(x^2+6x-9)/(x-2)-x=# #=lim_(x rarr2)(cancelx^2+6x-9-cancelx^2+2x)/(x-2)=# #=lim_(x rarr2)(8x-9)/(x-2)=8# So the equation of the slant asymptote is #y=x+8#

Benjamin Achtenberg · Answer

undefined To find the vertical asymptote of $ f(x) = \frac{x^2 + 6x - 9}{x - 2} $, determine the values of $ x $ that make the denominator equal to zero. In this case, the vertical asymptote occurs at $ x = 2 $.

To find the horizontal asymptote, compare the degrees of the numerator and denominator polynomials. Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote. Instead, there is an oblique (slant) asymptote.

To find the slant asymptote, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient obtained represents the equation of the slant asymptote. In this case, performing the division yields $ y = x + 8 $. Therefore, the slant asymptote is $ y = x + 8 $.