+71 What are the Holes of the following functions
1. f(x) = (x2 - 3x + 2)(x - 3) / (x - 1)(x2 - 5x + 6)
2. f(x) = (x2 - 9)(3x + 2) / (x2 - 4)(x - 3)
+248 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 
+130071 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)
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(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
