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

Slant asymptote: $y = \frac{3 x}{100}$
Vertical asymptote $x = - \frac{\sqrt[3]{2500}}{50}$
Horizontal asymptote:none

#### Explanation:

From the given: $y = \frac{3 {x}^{4} + 2 x + 1}{100 {x}^{3} + 2}$

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" " " " " " " " " " " " " " " " ")
$100 {x}^{3} + 0 \cdot {x}^{2} + 0 \cdot x + 2 \lceiling 3 {x}^{4} + 0 \cdot {x}^{3} + 0 \cdot {x}^{2} + 2 x + 1$
$\text{ " " " " " " " " " " " " " " " " " " " } \underline{3 {x}^{4} + 0 \cdot {x}^{3} + 0 \cdot {x}^{2} + \frac{3}{50} x}$
$\text{ " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " } \frac{97}{50} x + 1$

The result of the division will provide us the answers

part of the quotient $\frac{3 x}{100}$ will be the slant asymptote, that is

$y = \frac{3 x}{100}$

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

$100 {x}^{3} + 2 = 0$

$x = \sqrt[3]{- \frac{2}{100}}$

$x = - \frac{\sqrt[3]{2500}}{50}$

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

graph of slant asymptote $y = \frac{3 x}{100}$ and vertical asymptote
$x = - \frac{\sqrt[3]{2500}}{50}$

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

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