How do you identify all asymptotes or holes for #y=(4x^3+32)/(x+2)#?

1 Answer
Nov 3, 2016

Answer:

The function is a parabola

Explanation:

Let's do the long division
#color(white)(aaaa)##4x^3##color(white)(aaaaaaaaaaa)##+32##∣##x+2#
#color(white)(aaaa)##4x^3+8x^2##color(white)(aaaaaaaaa)##∣##4x^2-8x+16#
#color(white)(aaaaaa)##0-8x^2#
#color(white)(aaaaaaaa)##-8x^2-16x#
#color(white)(aaaaaaaa)##-8x^2-16x#
#color(white)(aaaaaaaaaaaa)##0+16x+32#
#color(white)(aaaaaaaaaaaaaa)##+16x+32#
#color(white)(aaaaaaaaaaaaaaaa)##+0+0#

The result is #(4x^3+32)/(x+2)=4x^2-8x+16#
#=4(x^2-2x+4)=4(x-2)^2#
This is a parabola
graph{4(x-2)^2 [-25.65, 25.67, -12.83, 12.84]}