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

1 Answer
Jan 1, 2017

Answer:

Horizontal: #larr y = 3/5 rarr#. Vertical: #uarrx = -0.7715135 and x = 0.5482532 darr#, nearly.

Explanation:

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

#f(x)=Q(x)+(R(x))/(S(x)) =3/5+(2x^2-3/5x+8/5)/(5x^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]} -