Question

Can anyone help?
f(x)=x^2-x-2/2x^2-x-21
I need to find hortizontal, vertical and oblique.
I couldn't find any vertical/oblique,
but for Hortizontal I have x-3 and x=7/2

Answer 1

degree of numerator is equal to the degree of the denominator so there is no oblique asymptote

Answer 2

but since they are equal the horizontal asymptote is the ratio of the leading coefficients, in your case it is \(y=\frac{1}{2}\)

Answer 3

how do i set the problem up?

Answer 4

i am not sure what you mean by "set the problem up"
it is a think problem
degree of numerator is 2
degree of denominator is 2
since they are equal, the horizontal asymptote is the ratio of the leading coefficients
the leading coefficient of the numerator is 1
the leading coefficient of the denominator is 2
the ratio is therefore \(\frac{1}{2}\) and your horizontal asymptote is \(y=\frac{1}{2}\)
you pretty much do it in your head

Answer 5

for vertical set \(2x^2-x-21=0\) and solve for \(x\)

Answer 6

which looks like what you did for "horizontal" but it should be vertical
you get
\[(x+3)(2x-7)=0\]
\[x=-3\] or
\[x=\frac{7}{2}\]for the vertical asymptotes

Answer 7

is my hortizontal correct?
I took (x-2)(x+1)/(x+3)(2x-7)

Answer 8

so my hortizontal is x=-3 and vertical is y=1/2