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

Dec 8, 2016

Horizontal: $\leftarrow y = o \rightarrow$
Vertical: $\uparrow x = a \in \left(- 3 , - 2\right) \mathmr{and} x = b \in \left(0 , 1\right) \downarrow .$ .

#### Explanation:

Let$f \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)}$

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 $x \to \pm \infty$.

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]}