How do I find the asymptote of this equation?

Question

jojokiw3 · Answer

$$y= \frac{ x^2 + 4x + 3 }{ x-1 }$$

jojokiw3 · Answer

I think I factor it first, which gets me: $$y = \frac{ (x+3)(x+1) }{ x-1 }$$ where my zeroes are 3 and 1.

zepdrix · Answer

Denominator is giving an asymptote.

zepdrix · Answer

Woops, note that your zeros are not 3 and 1.
$\large\rm (x+3)(x+1)=0\qquad\to\qquad (x+3)=0,\qquad (x+1)=0$
So your zeros are x=-3 and x=-1, ya?

zepdrix · Answer

In the land of math, you do not divide by 0.
We don't have a name for this number 3/0.
It simply doesn't exist, it is undefined.

Asymptote is that "bad fella",
he who shall not be named.
It's the location where we're dividing by 0.

So to find this location, we'll set our denominator equal to zero,
$\large\rm x-1=0$
and solve for x.

jojokiw3 · Answer

Haha I don't know how to find the slanted one. There's two asymptotes here.

jojokiw3 · Answer

You're back! I thought you left me in the inferno of this plane!

zepdrix · Answer

Sorry site is freezing on me >.<

zepdrix · Answer

Slant asymptote?
Apply long division or synthetic division,
the `quotient` gives you the equation of your oblique/slant asymptote.

zepdrix · Answer

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