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how do i find the oblique asymptots for f(x)=x^2-x-2/2x^2-x-21
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this does not have any oblique (slant) asymptotes.....
ok, that is what I thought. I also, got no vertical but for hortizontal I got y=-∞
there is a horizontal asymptote but it's not the one you wrote.... i'm checking now about the vertical....
it looks like you have two vertical asymptotes.
can you show me how?
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sure... this is your function... \[\huge y=f(x)=\frac{x^2-x-2}{2x^2-x-21}=\frac{(x-2)(x+1)}{(x+3)(2x-7)} \]
finding vertical asymptotes is easy.... all you need to do is factor the denominator and set it equal to zero. can you give me the x values that will make the denominator zero?
is the second part (x-2)(x+1)/(x+3)(2x-7)?
Vertical x-3,x=7/2 Hortizontal y=1/2
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