How do you find the vertical, horizontal and slant asymptotes of: #f(x)=1/(x^2-2x-8 )#?

Question

Serenity Johnson · Accepted Answer

The vertical asymptotes are #x=-2# and #x=4#
The horizontal asymptote is #y=0#
No slant asymptote.

Let's factorise the denominator #x^2-2x-8=(x+2)(x-4)# So, #f(x)=1/(x^2-2x-8)=1/((x+2)(x-4))# As you cannot divide by #0#, #x!=-2# and #x!=4# The vertical asymptotes are #x=-2# and #x=4# As the degree of the numerator is #<# than the degree of the denominator, there are no slant asymptote. #lim_(x->+-oo)f(x)=lim_(x->+-oo)1/x^2=0# The horizontal asymptote is #y=0# graph{(y-1/(x^2-2x-8))(y)(y-50x-100)(y-50x+200)=0 [-9.84, 10.18, -5.175, 4.825]}

Lincoln Anderson · Answer

undefined Vertical asymptotes occur at the zeros of the denominator, so set the denominator equal to zero and solve for x: x^2 - 2x - 8 = 0. Factoring gives (x - 4)(x + 2) = 0. Therefore, the vertical asymptotes are x = 4 and x = -2.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (0) is less than the degree of the denominator (2), the horizontal asymptote is y = 0.

To find the slant asymptote, perform polynomial long division or use partial fractions to divide the numerator by the denominator. The result will be the equation of the slant asymptote, which is y = 0x + 1/1 = 1.