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

1 Answer
Feb 18, 2018

Answer:

See below.

Explanation:

Start by simplifying the expression:

#(x^4-x^2)/(x(x-1)(x+2))#

#(x^2(x^2-1))/(x(x-1)(x+2))#

#(x^2-1)=(x+1)(x-1)# ( Difference of two squares )

#(x^2(x+1)(x-1))/(x(x-1)(x+2))#

#(x^cancel(2)(x+1)cancel((x-1)))/(cancelxcancel((x-1))(x+2))#

#(x(x+1))/(x+2)=(x^2+x)/(x+2)#

Vertical asymptotes occur where the function is undefined. This can be seen to be when #x=-2# (division by zero)

So the line #x=-2# is a vertical asymptote.

We now divide:

#(x^2+x)/(x+2)#

The quotient of this will be the oblique asymptote:

#(x+2)|bar(x^2+x)#

#color(white)(888888888)x#
#(x+2)|bar(x^2+x)#

#color(white)(888888888)x-1#
#(x+2)|bar(x^2+x)#
#color(white)(88888888)x^2+2x#
#color(white)(88888888888)-x#
#color(white)(88888888888)-x-2#
#color(white)(88888888888.888888)2#

We do not need to be concerned with the remainder. So the oblique asymptote is the line:

#y=x-1#

This is the line #(x^2+x)/(x+2)# approaches as #x->+-oo#

Graph:

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