# How do you find vertical, horizontal and oblique asymptotes for  f(x)= (2x^3+11x^2+5x-1)/(x^2+6x+5)?

Dec 27, 2016

Vertical : $x = - 1 \mathmr{and} x = - 5$.
Slant: $y = 2 x - 1$.
See Socratic graph for the curve and the asymptotes.

#### Explanation:

y = f(x)=2x-1+(x-4)/((x+1)(x+5)

y = quotient = $2 x - 1$ gives the slant asymptote, with slope 2.

The factors ( x + 1 ) and ( x + 5 ) in the denominator of the remainder

equated to 0 give the vertical asymptotes.

The ad hoc graphs are not in uniform as-is scales.

They serve the purpose of revealing asymptoticity of the three

asymptotes, with respect to the three branches of the graphs, taken

in pairs..

graph{(y(x+1)(x+5)-2x^3-11x^2-5x+1)(x+5)(y-2x+1)(x+1)=0 [-10, 10, -30, `30]}

graph{(y(x+1)(x+5)-2x^3-11x^2-5x+1)(x+5)(y-2x+1)(x+1)=0 [-40, 40, -82.5, 57.5]}