1. ## Curve Sketching

Given f(x)=(x^2-x-2)/(x^2+x), find the vertical and horizontal asymptotes and sketch the graph near the asymptotes.

VA
x=0 or x=-1

HA
as x -> +-infinity, y->1

So that is my work above and I graphed this all too. But.... my question comes here.
When I graph the function f(x) using software, there appears to be no vertical asmptote of x=-1, only x=0 is a VA... This is really strange because I do believe there are two asymptotes..

to find asymptote, set denominator =0 and solve for x.
x^2 + x = 0
x(x+1)=0
x=0 or x=-1
so?? what did I do wrong now?

Note: I did not type the eqtn of the function wrong into graphing software.

2. Originally Posted by skeske1234
Given f(x)=(x^2-x-2)/(x^2+x), find the vertical and horizontal asymptotes and sketch the graph near the asymptotes.

VA
x=0 or x=-1

HA
as x -> +-infinity, y->1

So that is my work above and I graphed this all too. But.... my question comes here.
When I graph the function f(x) using software, there appears to be no vertical asmptote of x=-1, only x=0 is a VA... This is really strange because I do believe there are two asymptotes..

to find asymptote, set denominator =0 and solve for x.
x^2 + x = 0
x(x+1)=0
x=0 or x=-1
so?? what did I do wrong now?

Note: I did not type the eqtn of the function wrong into graphing software.
You did nothing "wrong". You overlooked something...

Note that $\displaystyle \frac{x^2-x-2}{x^2+x}=\frac{\left(x-2\right){\color{red}\left(x+1\right)}}{x{\color{re d}\left(x+1\right)}}$

Since the term $\displaystyle x+1$ appears in both the numerator and the denominator, there is a hole in the graph at $\displaystyle x=-1$, not a vertical asymptote.