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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)

 May 19, 2019
 #1
avatar+224 
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  smiley

 May 19, 2019
 #2
avatar+111389 
+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

 

 

 

cool cool cool

 May 19, 2019

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