How do you find the slant asymptote of $y=((4x^3)+(x^2)+x+4)/((x^2)+5x)$ ?

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Answer 1 · 2016-02-20T04:16:21

Slant asymptote is $y=(4x-19)$

While vertical asymptotes are given by solution of $(x^2+5x)=0$ i.e. $x(x+5)=0$ i.e. they are $x=0$ and $x=-5$

To find slant asymptote of $y=(4x^3+x^2+x+4)/(x^2+5x)$ , divide $(4x^3+x^2+x+4)$ by $(x^2+5x)$ , o

$4x(x^2+5x)-19(x^2+5x)+96x+4$ i.e.

$y=(4x-19)+(96x+4)/(x^2+5x)$

Hence slant asymptote is $y=(4x-19)$

How do you find the slant asymptote of y=((4x^3)+(x^2)+x+4)/((x^2)+5x)y=((4x^3)+(x^2)+x+4)/((x^2)+5x)?