How did (2,4) become a point in the graph of f(x)=(x^2-4)/(x-2)?
Stacey Warren - Expert brainly.com
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Strictly speaking (2,4) is not part of f(x) as defined.
f(x)= x+2, for x \(\ne\) 2
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)
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this is the graph my professor drew @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
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?
no, not a random point.
if you have time , see