how do i find the oblique asymptots for
f(x)=x^2-x-2/2x^2-x-21

Question

how do i find the oblique asymptots for
f(x)=x^2-x-2/2x^2-x-21

anonymous · Answer

this does not have any oblique (slant) asymptotes.....

anonymous · Answer

ok, that is what I thought. I also, got no vertical but for hortizontal I got y=-∞

anonymous · Answer

there is a horizontal asymptote but it's not the one you wrote....
i'm checking now about the vertical....

anonymous · Answer

it looks like you have two vertical asymptotes.

anonymous · Answer

can you show me how?

anonymous · Answer

sure...
this is your function...
$$\huge y=f(x)=\frac{x^2-x-2}{2x^2-x-21}=\frac{(x-2)(x+1)}{(x+3)(2x-7)} $$

anonymous · Answer

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?

anonymous · Answer

is the second part (x-2)(x+1)/(x+3)(2x-7)?

anonymous · Answer

Vertical x-3,x=7/2
Hortizontal y=1/2