since the denominator is (x+1)(x+2) you'd expect to see vertical asymptotes at x= -1 and -2
But the numerator is (x-1)(x+2)(x-3) so at everywhere except x = -1,-2
f(x) = (x-1)(x+2)(x-3) / (x+1)(x+2) = (x-2)(x-3) / (x+1)
= x-5 + 8/(x+1)
So, there is a slant asymptote of y=x-5
So, the only vertical asymptote is at x = -2. There is a hole at x = -1, since f(-1) = 0/0 is undefined.
There is no horizontal asymptote since the numerator has greater degree than the denominator.
For x ≠ -1, -2
f(x) =
Given the function
f(x)=x^3-2x^2-5x+6/x^2+3x+2
a) Determine algebraically the equation for any asymptotes (HA, VA and or OA)
b)State the domain of the function
c)State the behaviour of the function at each of the asymptotes
3 answers
sorry - lost track of where I was on the screen. But, the info is there/
its ok, i just didn't know how to verify if there was a horizontal asymptote and I didn't know that x=-1 is hole, thank you for the help!
1 answer