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If \(\frac{3x^2-4x+1}{x-1}=m\), and x can be any real number except 1, what real value(s) can m NOT have? Enter all the values, separated by commas.

 Aug 22, 2022
 #1
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First, I would factor the quadratic in the numerator

 

3x^2-4x+1 can be factored into (3x-1)(x-1)

 

So you get (3x-1)(x-1)/(x-1)

 

Notice how you can "cancel out" the x-1, you can't necessarily cancel it out, but it does become a removable discontinuity, meaning that the graph of the function looks exactly like the graph of y=3x-1, but when x is 1, there's a hole in the graph because the function is not defined at that point (since plugging in a 1 in the denominator will still cause the denominator to equal 0, and as you know, 0s in denominators are illegal in math, but it's not completely illegal in this equation because we can temporarily cancel the x-1 term out)

 

This is what the equation looks like in desmos

 

 

So this means that the function is defined everywhere except when x is equal to 1

 

Thus, plug in 1 for the equation y=3x-1

 

f(1) ≠ 3(1) - 1

 

f(1) ≠ 2

 

So m ≠ 2

 Aug 22, 2022
edited by mimi997  Aug 22, 2022

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