Asked by johnny
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
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
Answers
Answered by
Steve
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) =
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) =
Answered by
Steve
sorry - lost track of where I was on the screen. But, the info is there/
Answered by
johnny
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!
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