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 = 0xv=0 and n = m+1n=m+1 with n = 2n=2 and m = 1m=1
Slant asymptotes are obtained considering (p_n(x))/(p_{n-1}(x))
approx y = a x+b pn(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^2-2x-3)/(-4x)pn(x)pn−1(x)=x2−2x−3−4x
p_n(x) = p_{n-1}(x)(a x+b)+r_{n-2}(x)pn(x)=pn−1(x)(ax+b)+rn−2(x)
x^2-2x-3 =(-4x)(a x + b) + cx2−2x−3=(−4x)(ax+b)+c
equating coefficients
{
(-3 - c=0),
(-2 + 4 b=0),
(1 + 4 a=0)
:}
solving for a,b,c we get {a = -(1/4), b = 1/2, c = -3}
and substituting
y = -x/4+1/2
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