Find all Vertical Asymptotes.
The answer is x=1;x=-2, but I can't figure out how. The x^3 in the denominator threw everything off.

Question

Find all Vertical Asymptotes.
The answer is x=1;x=-2, but I can't figure out how. The x^3 in the denominator threw everything off.

anonymous · Answer

$$h(x)=\frac{ x^3+x^2+x+1 }{ x^3-3x+2 }$$

anonymous · Answer

To find vertical asymptotes you are generally concerned with the denominator. I like to factor the denominator to make the job easier:$$x ^{2}-3x+2=(x-1)(x-2)$$
Now set that equal to zero and you obtain that the function is undefined (a vertical asymptote) when x=1 and x-2.

anonymous · Answer

That should say x=2 not x-2. Sorry typo ;p

anonymous · Answer

However, in the denominator its $$x^3$$not$$x^2$$

anonymous · Answer

Oh I'm sorry I thought that was a 2. You're going to make me work a little harder to factor that lol. Just a sec.

anonymous · Answer

So the correct denominator is $$x ^{3}-3x+2=(x+2)(x-1)^{2}=0$$ From this we can tell that the roots are x=-2 and x=1.

anonymous · Answer

Oh so that's how you factor it.

anonymous · Answer

Do you know how to use synthetic division? I prefer that method when dividing large polynomials by potential factors.

anonymous · Answer

I said large. What I meant was "any." lol ;p

anonymous · Answer

I see. Well Thank you for helping me Jonathan.

anonymous · Answer

Np good luck with your classes.

anonymous · Answer

Thank you.