How do you find the vertical, horizontal or slant asymptotes for #f(x) = ( 3x^5 + 1) /( 2x^6 + 3x -1)#?

1 Answer
Dec 8, 2016

Answer:

Horizontal: #larr y = o rarr#
Vertical: #uarr x=a in (-3, -2) and x = b in (0, 1) darr.# .

Explanation:

Let# f(x) = (N(x))/(D(x))#

The number of changes in the sigs of the coefficients in #D(x)= and

D(-x) are both 1 only. So, the number of real zeros of D(x) is either 2

or 0. Easily D(0)D(1) and D(-2)D(-1) are both negative. So, two real

roots x = a and x = b are bracketed in (-2, -1) and (0, 1). And so, the

vertical asymptotes are x = a and x = b.

#y=(3+1/x^5)/(x(2+3/x^5-1/x^6) to 0#, as #x to +-oo#.

So, y = 0 is the horizontal asymptote.

The x-intercept (y = 0 ) is -(-1/3)^(1/5)=-0.80, nearly.

y-intercept ( x = 0 ) is -1.

graph{y(2x^6+3x-1)-3x^5-1=0.2 [-10, 10, -5, 5]}