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anonymous
 one year ago
How did (2,4) become a point in the graph of f(x)=(x^24)/(x2)?
anonymous
 one year ago
How did (2,4) become a point in the graph of f(x)=(x^24)/(x2)?

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phi
 one year ago
Best ResponseYou've already chosen the best response.1Strictly speaking (2,4) is not part of f(x) as defined. f(x)= x+2, for x \(\ne\) 2

phi
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434548565383:dw this is the graph my professor drew @phi

phi
 one year ago
Best ResponseYou've already chosen the best response.1ok, 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
 one year ago
Best ResponseYou've already chosen the best response.1now 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 (x2) when x is exactly 2 that becomes (22) = 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi oh so was it just a random point? just to show the continuity of the line on the graph?

phi
 one year ago
Best ResponseYou've already chosen the best response.1no, not a random point. if you have time , see https://www.khanacademy.org/math/algebra2/polynomial_and_rational/simplifyingrationalexpressions/v/simplifyingrationalexpressionsintroduction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi thanks for your help!
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