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