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
degree of numerator is equal to the degree of the denominator so there is no oblique asymptote
but since they are equal the horizontal asymptote is the ratio of the leading coefficients, in your case it is \(y=\frac{1}{2}\)
how do i set the problem up?
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
for vertical set \(2x^2-x-21=0\) and solve for \(x\)
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
is my hortizontal correct? I took (x-2)(x+1)/(x+3)(2x-7)
so my hortizontal is x=-3 and vertical is y=1/2
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