In a polynomial fraction f(x) = (p_n(x))/(p_m(x))f(x)=pn(x)pm(x) we have:
1)1) vertical asymptotes for x_vxv such that p_m(x_v)=0pm(xv)=0
2)2) horizontal asymptotes when n le mn≤m
3)3) slant asymptotes when n = m + 1n=m+1
In the present case we have x_v = {-2, 2}xv={−2,2} and n = m+1n=m+1 with n = 3n=3 and m = 2m=2
Slant asymptotes are obtained considering (p_n(x))/(p_{n-1}(x))
approx y = a x+bpn(x)pn−1(x)≈y=ax+b for large values of abs(x)|x|
In the present case we have
(p_n(x))/(p_{n-1}(x)) = x^3/(x^2-4)pn(x)pn−1(x)=x3x2−4
p_n(x)=p_{n-1}(x)(a x+b)+r_{n-2}(x)pn(x)=pn−1(x)(ax+b)+rn−2(x)
r_{n-2}(x)=c x + drn−2(x)=cx+d
x^3 = (x^2-4)(a x + b) + c x + dx3=(x2−4)(ax+b)+cx+d
equating coefficients
{
(4 b - d=0), (4 a - c=0), (-b=0), (1 - a=0)
:}
solving for a,b,c,d we have {a = 1, b = 0, c = 4, d = 0}
substituting in y = a x + b
y = x
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