# How do you find vertical, horizontal and oblique asymptotes for #f(x)= (x^3+8) /(x^2+9)#?

The oblique asymptote is

No vertical and horizontal asymptotes

So there are no vertical asymptotes.

Let's do a long division

So,

graph{(y-(x^3+8)/(x^2+9))(y-x)=0 [-5.546, 5.55, -2.773, 2.774]}

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To find the vertical asymptotes, set the denominator equal to zero and solve for ( x ). In this case, for ( f(x) = \frac{x^3 + 8}{x^2 + 9} ), the denominator ( x^2 + 9 ) equals zero when ( x^2 = -9 ). Since ( x^2 ) cannot be negative for real ( x ), there are no vertical asymptotes.

For horizontal asymptotes, compare the degrees of the numerator and the denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at ( y = 0 ). If the degree of the numerator equals the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. In this case, the degrees of the numerator and denominator are 3 and 2 respectively. Therefore, there are no horizontal asymptotes.

For oblique asymptotes, if the degree of the numerator is exactly one more than the degree of the denominator, then the oblique asymptote can be found by performing polynomial long division. However, in this case, the degree of the numerator is not exactly one more than the degree of the denominator. Therefore, there are no oblique asymptotes for this rational function.

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