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 = -2xv=−2 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+b xpn(x)pn−1(x)≈y=a+bx for large values of abs(x)|x|
In the present case we have
(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4)/(x + 2) pn(x)pn−1(x)=2x2−3x+4x+2
p_n(x) = p_{n-1}(x)(a x + b) + r_{n-2}(x)pn(x)=pn−1(x)(ax+b)+rn−2(x)
(p_n(x))/(p_{n-1}(x)) = (2x^2-3x+4) =(x+2) (a x +b)+cpn(x)pn−1(x)=(2x2−3x+4)=(x+2)(ax+b)+c
equating we have
{
(4 - 2 b - c=0), (-3 - 2 a - b=0), (2 - a=0)
:}
Solving for a,b,c we have a = 2,b=-7,c=18
so the slant asymptote reads
y = 2x-7
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