**What are the Holes of the following functions**

**1. f(x) = (x ^{2} - 3x + 2)(x - 3) / (x - 1)(x^{2} - 5x + 6)**

**2. f(x) = (x ^{2} - 9)(3x + 2) / (x^{2} - 4)(x - 3)**

Mathrules May 19, 2019

#1**0 **

I'm assuming by "holes" you mean asymptotes, or values that make the function undefined.

To solve a problem like this, you would start by factoring the numerator and denominator completelty.

\(f(x)=\frac{(x^2 - 3x + 2)(x - 3)}{ (x - 1)(x^2 - 5x + 6)}=\frac{(x-2)(x-1)(x-3)}{(x-1)(x-3)(x-2)}=0\)

In this case, all the factors cancel out and you are left with \(f(x)=0\)

But by simplifing you made the function work for every number, when there are numbers that won't work in the original equation.

If x=1, x=2 or x=3 the denominator is 0 and dividing by 0 is undefined.

Therefore the "holes" or asymptotes of the function are at \(\boxed{x=1, x=2, x=3}\)

I hope this helps and you can figure out the second problem from my solution

power27 May 19, 2019

#2**+1 **

2. Factor both numerator / denominator

( x^2 - 9) factors as ( x + 3) ( x - 3)

(x^2 - 4) factors as ( x + 2) ( x - 2)

So we have

( x + 3) ( x - 3) (3x + 2)

__________________

(x + 2) ( x - 2) ( x - 3)

Notice that (x - 3) "cancels" in both the numerator and denominator

Setting this to 0, we have that x - 3 = 0 ⇒ x = 3

So....we have a " hole" at x = 3

And vertical asymptotes at x = -2 and x = 2

Here's a graph : https://www.desmos.com/calculator/g9nniltjdu

CPhill May 19, 2019