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The numerator is not factorable, but graphing shows one root near 1.25. So this will be a root of the function f(x)...not that you needed that!
For vertical asymptotes, the denominator must equal zero. This cannot happen (except with complex numbers), so there aren't any v.a.
Horizontal and oblique asymptotes require us to look at end behavior. Since we have a cubic divided by a quadratic, the function f(x) will have linear end behavior. We just have to find the equation of the line.
Polynomial division (see link) gives `f(x)=5-3 x-7/(4+x^2)` , which will behave like y = -3x + 5 when x gets very large (because `7/(4+x^2)`
will approach zero). This is your oblique asymptote!
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