How do you find vertical, horizontal and oblique asymptotes for #[(3x^2) + 14x + 4] / [x+2]#?

1 Answer
Dec 27, 2016

Answer:

Vertical: #uarr x=-2 darr#
Oblique: #y=3x+8#, in both directions. See the graphs.

Explanation:

#(y(x+2)-3x^2) -14x-4=0#

Reorganizing to the form (ax+bx+c((a'x+b/y+c')=k,

#(x+2)(y-3x-8)=-12#

This represents a hyperbola with asymptotes

#(x+2)(y-3x-8)#=0

The first graph is asymptotes inclusive and the second is for the

hyperbola, sans asynptotes.

Note: The second degree equation

#ax^2+2hxy+by^2+...=0# represents a hyperbola, if #ab-h^2 < 0#. It

is so for the given equation.

graph{(y-3x-8)(x+2)((y-3x-8)(x+2)+14)=02 [-80, 80, -40, 40]}

graph{(y-3x-8)(x+2)+14=0 [-80, 80, -40, 40]}