# How do you find the asymptotes for f(x)=(3x^4 + 2x^2 + 1 )/ (5x^4 + x -1)?

Jan 1, 2017

Horizontal: $\leftarrow y = \frac{3}{5} \rightarrow$. Vertical: $\uparrow x = - 0.7715135 \mathmr{and} x = 0.5482532 \downarrow$, nearly.

#### Explanation:

graph{5x^4+x-1 [-0.884, 0.8805, -0.443, 0.439]} By actual division,

$f \left(x\right) = Q \left(x\right) + \frac{R \left(x\right)}{S \left(x\right)} = \frac{3}{5} + \frac{2 {x}^{2} - \frac{3}{5} x + \frac{8}{5}}{5 {x}^{4} + x - 1}$

y = Q and S(x) = 0 give the asymptotes

The number of changes in signs of the coefficients in S(x) are each 1.

So, the number of real zeros of S is either 1+1 = 2 or none. The first

graph for S reveals two zeros of S = 0, near -0.77 and 0.548, nearly.

I have improved these approximations to 7-sd given in the answer,

using an iterative numerical method.

The second graph is for asymptotes-inclusive f(x).

graph{(x+.771)(x-.548)(y-3/5)(y-(3x^4+2x^2+1)/(5x^4+x-1))=0 [-14.12, 14.11, -7.06, 7.05]} -