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What are the points of discontinuity? Are they all removable? y=(x-3)/x^2-12x+27
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\[y=\frac{ (x-3) }{ x ^{2} -12x+27 }\]
Choices:
Factor the denominator and look at what cancels (if anything).
That leaves x-9 in the denominator.
That which makes the cancelled binomial = 0 represents a "hole" or point of discontinuity. Normally the zeroes of the denominator are vertical asymptotes, but when a factor of the denominator cancels you get holes instead.
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The answer must be x=9, x=3; no
Correct?
The graph is discontinuous at 9 and 3, but only 3 is removable. At x=3 there is a hole but the function would be continuous to that point and just after. At x=9 the graph is totally disconnected separated by an asymptote.
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