q(x) = x^4 + 3x^3 - 13x^2 - 51 - 36 / x^3 + 9x^2 + 23x + 15 = (x + 3)^2(x - 4)(x + 1) / (x + 3)(x + 5)(x + 1)
Horizontal or oblique asymptotic equation
since
q(x) = ((x + 3)^2(x - 4)(x + 1)) / ((x + 3)(x + 5)(x + 1))
= x-6 + 18/(x+5)
the slant asymptote is y = x-6
there is no horizontal asymptote
The vertical asymptote is at x = -5
Apologies for the oversight in the calculation of the slant asymptote in the previous response. You are correct. The correct slant asymptote for the rational function q(x) is y = x - 6.
Given the function:
q(x) = ((x + 3)^2(x - 4)(x + 1)) / ((x + 3)(x + 5)(x + 1))
The slant (oblique) asymptote is indeed y = x - 6.
Thank you for pointing out the error, and I appreciate your correction.
The horizontal asymptote of a rational function is given by comparing the degrees of the numerator and the denominator. In this case, the degrees of the numerator and denominator are the same (4), so the horizontal asymptote is given by the ratio of the leading coefficients:
Horizontal asymptote: y = x^4 / x^3 = x
To find the oblique asymptote, we need to perform long division of the numerator by the denominator. The result of the division is:
(x^4 + 3x^3 - 13x^2 - 51) / (x^3 + 9x^2 + 23x + 15) = x + 1
Therefore, the oblique asymptote is given by the equation y = x + 1.