How do you find vertical, horizontal and oblique asymptotes for $(x^4 - 2x + 3) / (6 - 5x^3)$ ?

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Answer 1 · 2017-02-18T11:43:04

Slant asymptote : $x+5y=0$ .
Vertical asymptote : $uarrx = (6/5)^(1/3)=1.063darr$ , hearly.

See depiction by Socratic graphs.

graph{(x^4-2x+3)/(6-5x^3) [-10, 10, -5, 5]}
graph{(x-1-.025y)((x^4-2x+3)/(6-5x^3)-y)(x+5y)=0 [-20, 20, -10, 10]}

By actual division,

$y = -x/5+(3-4/5x) / (5(((6/5)^(1/3))^3-x^3)$

$=x/5+(3-4/5x)/(5(((6/5)^(1/3)-x)((6/5)^(2/3)+(6/5)^(1/3)x+x^2)))$

revealing the slant asymptote

$y =-x/5$ and the vertical asymptote

$x=(6/5)^(1/3)=1.063$ , nearly

How do you find vertical, horizontal and oblique asymptotes for (x^4 - 2x + 3) / (6 - 5x^3)(x^4 - 2x + 3) / (6 - 5x^3)?