How do you find the vertical, horizontal or slant asymptotes for #f(x) = (3x^4 + 2x +1 )/ (100x^3 + 2)#?

1 Answer

Answer:

Slant asymptote: #y=(3x)/100#
Vertical asymptote #x=-root3(2500)/50#
Horizontal asymptote:none

Explanation:

From the given: #y=(3x^4+2x+1)/(100x^3+2)#

To obtain the slant asymptote, we divide numerator by the denominator and take note of the quotient. We can divide because the dividend is higher in degree.

By long division

#" " " " " " " " " " " " " " " " " " " "underline(" " "3x/100" " " " " " " " " " " " " " " " ")#
#100x^3+0*x^2+0*x+2 |~3x^4+0*x^3+0*x^2+2x+1#
#" " " " " " " " " " " " " " " " " " " " "underline(3x^4+0*x^3+0*x^2+3/50x)#
#" " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " "97/50x+1#

The result of the division will provide us the answers

part of the quotient #(3x)/100# will be the slant asymptote, that is

#y=(3x)/100#

To obtain the Vertical Asymptote
Equate the divisor to zero then solve for x, that is

#100x^3+2=0#

#x=root3(-2/100)#

#x=-root3(2500)/50#

graph of #y=(3x^4+2x+1)/(100x^3+2)#
graph{(y-(3x^4+2x+1)/(100x^3+2))=0[-20,20,-10,10]}

graph of slant asymptote #y=(3x)/100# and vertical asymptote
#x=-root3(2500)/50#

graph{(y-(3x)/100)( y+1000000x+1000000root3(2500)/50 )=0[-20,20,-10,10]}

God bless....I hope the solution is useful