The graph of the function either touches or passes through the x axis where x equals 2.
It is asymptotic (approaches, but never quite equals) to the line x=3 (that is, it points nearly straight up where x equals 3).
This means that (a) it must, on average, slope upward between where x equals 2 and x equals 3, and (b) it can't be a straight line - it has to either (a) be curved or (b) be discontinuous (have a break in the graph somewhere between x equals 2 and x equals 3), to do this.
2006-10-24 07:34:32
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answer #1
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answered by Anonymous
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1) when x=2, y=0. x=2 is a root of f(x), i.e. a point where the graph intersects the x-axis (y=0 is the x-axis)
2) as x approaches 3 the limit of f(x) is infinity. at x=3, the value of f(x) is undefined. but as you "ride" along the graph and get closer and closer to x=3, the value of the function gets larger and larger toward infinity. The minus sign to the right of the 3 in the limit expression means the limit is only being evaluated as you approach x=3 from values less than x=3. the behavior may be different as you approach it from the other side (for example it could be negative infinity).
2006-10-24 07:31:06
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answer #2
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answered by T M 6
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(1) the point (2,0) lies on the graph of f. Also, 2 is an x-intercept of the graph, meaning the graph intercepts the x-axis at x=2.
(2) as x approaches 3 from the left, f(x) tends toward positive infinity, so f has a vertical asymptote at x=3. This means that the graph of f approaches the vertical line x=3, from the left side.
2006-10-24 07:27:42
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answer #3
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answered by James L 5
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If f(2)=0, then there is a point (2,0) on the curve.
the curve increases and as the value of X gets close to 3, the value gets increasingly large, until at 3, the function is undefined.
If f(x) = -1-[1/x-3], there is a point (2,0) and the value at 3 is undefined.
2006-10-24 08:15:40
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answer #4
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answered by davidosterberg1 6
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