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

Question

Sophie Adams · Accepted Answer

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

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#

Chloe Alexander · Answer

undefined To identify all asymptotes or holes for $ g(x) = \frac{x^3 + 3x^2 - 16x + 12}{x - 2} $, we first look for any vertical asymptotes by finding the values of $ x $ that make the denominator zero, since division by zero is undefined.

Setting the denominator $ x - 2 $ equal to zero gives $ x = 2 $. Therefore, there is a vertical asymptote at $ x = 2 $.

Next, we check for any holes in the graph by simplifying the function to see if there are any common factors in the numerator and denominator that can be canceled out.

Factorizing the numerator $ x^3 + 3x^2 - 16x + 12 $ gives $ (x - 1)(x - 2)(x + 6) $.

Since $ (x - 2) $ is a factor of both the numerator and the denominator, it cancels out, leaving us with $ x^2 + 5x - 6 $.

This simplifies to $ (x + 6)(x - 1) $.

Therefore, there is a hole at $ x = 2 $ because $ (x - 2) $ cancels out from the numerator and the denominator, leaving a hole at that point.