How do you find the vertical, horizontal or slant asymptotes for # (5x^3-2x^2+1)/(4x^3+2x-7)#?

1 Answer
Jan 6, 2017

Answer:

Vertical : #uarr x = 1. 067 darr#, nearly.
Horizontal : #larr y = 1.25 rarr#

Explanation:

By actual division,

the given function

#y =5/4+(-2x^2-5x/2+35/4)/(4x^3+2x-7)#, giving

an asymptote y = quotient =5/4.

After due analysis of the second graph of the denominator-cubic,

for its zeros,

there is just another asymptote

x = 1.067, nearly.

See the asymptotes-inclusive first graph for the given function.

graph{(y-5/4)(x-1.067)(y-(5x^3-2x^2+1)/(4x^3+2x-7))=0x^2 [-5, 5, -2.5, 2.5]}

graph{4x^3+2x-7 [-20, 20, -10, 10]}