How did (2,4) become a point in the graph of f(x)=(x^2-4)/(x-2)?

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How did (2,4) become a point in the graph of f(x)=(x^2-4)/(x-2)?

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  • phi
Strictly speaking (2,4) is not part of f(x) as defined. f(x)= x+2, for x \(\ne\) 2
  • phi
Yes, but "they" did it incorrectly. It this is in a book or question, I would ask the teacher about it... because (2,4) is not part of f(x)
  • phi
*IF this is...

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|dw:1434548565383:dw| this is the graph my professor drew @phi
  • phi
ok, notice the circle at (2,4) is *open* that is how people show that a point is *not* on the line. at exactly (2,4) there is a "hole" in the line
  • phi
now the question is, "why is there a hole?" the answer: if you look at the original definition of f(x) you see you divide by (x-2) when x is exactly 2 that becomes (2-2) = 0 and we do not allow divide by 0 (the quotient is undefined) so we put an open circle at (2,4) to show that point is not on the line.
@phi oh so was it just a random point? just to show the continuity of the line on the graph?
  • phi
no, not a random point. if you have time , see https://www.khanacademy.org/math/algebra2/polynomial_and_rational/simplifying-rational-expressions/v/simplifying-rational-expressions-introduction
@phi thanks for your help!

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