How do you identify all asymptotes or holes for #g(x)=(x^3+3x^2-16x+12)/(x-2)#?

1 Answer
Oct 22, 2016

Answer:

The numerator is divisible by the denominator
#g(x)=(x+6)(x-1)#
There are no asymptotes or holes

Explanation:

Let's do a long division

#x^3+3x^2-16x+12##color(white)(aaa)##∣##x-2#
#x^3-2x^2##color(white)(aaaaaaaaaaaaa)##∣# #x^2+5x-6#
#0+5x^2-16x#
#color(white)(aaaaa)##0-10x#
#color(white)(aaaaaaaa)##-6x+12#
#color(white)(aaaaaaaaaaa)##0+0#

So #g(x)=(x^3+3x^2-16x+12)/(x-2)=x^2+5x-6#

We can factorise so #g(x)=g(x)=(x+6)(x-1)#

This is a parabola
there are no asymptotes or holes

#g(x)# is defined on #RR#