How do you find the vertical, horizontal and slant asymptotes of #b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)#?

1 Answer
Jan 21, 2017

Answer:

Vertical asymptotes : # uarr x = 0.861 darr and uarr x = -1.227 darr#

Explanation:

The vertical asymptotes are given by

x = zero of #d(x)=2x^4+x^3-x^2-1#

By considering the chanes 1 and 1 in the signs of the coefficients of

d(x), d has either two real roots are none,

The first graph for d reveals two real zeros, near 0.8 and -1.2.

The second near- linear mapping near x = 0,8 reveals this zero as

0.861, nearly, and likewise, the third graph reveals the second zero

near-1.227.

graph{2x^4+x^3-x^2-1 [-1.25, 1.25, -0.625, 0.625]}

graph{2x^4+x^3-x^2-1 [.86, .87, -.43.43]}

graph{2x^4+x^3-x^2-1 [-1.228, -1.226, -.43.43]}