Question

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

Answer 1

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