I need some help graphing this. (2x^2+2x-4)/(x^2+x).
I have to show the slant asymptote and vertical asymptotes.
I found that the zeros are 1 and -2, the horizontal asymptote is y=2.
What I don't know is the vertical asymptotes. I think the are at x=-1 and 0 because that is what makes the denominator=0. And I think I know that as the limit as x approaches -1 from both sides, it appoaches negative infinity, so the graph goes down. I am getting confused on whether the x=0 is also a VA and if so how to deal with it.

Question

I need some help graphing this. (2x^2+2x-4)/(x^2+x).
I have to show the slant asymptote and vertical asymptotes. 
I found that the zeros are 1 and -2, the horizontal asymptote is y=2.
What I don't know is the vertical asymptotes. I think the are at x=-1 and 0 because that is what makes the denominator=0. And I think I know that as the limit as x approaches -1 from both sides, it appoaches negative infinity, so the graph goes down. I am getting confused on whether the x=0 is also a VA and if so how to deal with it.

ranga · Answer

$\Large \frac{2x^2+2x-4}{x^2+x} = \frac{2(x^2+x-2)}{x(x+1)} = \frac{2((x+2)(x-1)}{x(x+1)}$ 
Zeros at x = -2 and x = 1.
Vertical asymptotes at x = 0 and x = -1.

anonymous · Answer

Thanks!  I think I had it... just got a little mixed up when I did one of the limits.  Makes sense now though...

ranga · Answer

No slant asymptotes here because the degree of the numerator is NOT one more than the denominator.
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