Thread: 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.

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

Since the term appears in both the numerator and the denominator, there is a hole in the graph at , not a vertical asymptote.