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Precalculus Mathematics Homework Help
How to determine a hole in a graph?
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[QUOTE="Stephen Tashi, post: 4980039, member: 186655"] It's hard to express your idea with mathematical precision. To put your question in a sophisticated way: When does a function that is the ratio of polynomial factors have a "removable singularity"? You want to know when an otherwise solid graph of a function has a hole at a (finite) point (x,y). One can define funtions in complicated ways using if...then rules as well as algebraic expressions. Suppose we only consider a function defined by the ratio of polynomial factors. (i.e. its is a single fraction , not a sum of several fractions). Functions like [itex] f(x) =\frac{(x^2 + 1) (2x + 3)} { (x^2 + 1)} [/itex] don't have a hole in their graph because (in the real number system) there is no value of [itex] x [/itex] that would make the denominator zero. So the fact that the numerator and denominator have a common factor does not always imply the function has a hole in its graph. Looking for terms that cancel out, doesn't automatically locate a hole in the graph. By contrast, the graph of the function [itex] g(x) = \frac{(x^2 -1 )(2x + 3)}{(x^2 - 1) } [/itex] has a holes when [itex] x = 1 [/itex] and [itex] x = -1 [/itex]. The graph of the function [itex] h(x) = \frac{ 6 + (x^2 )}{(x^2-1) } [/itex] doesn't exist at the values [itex] x = 1 [/itex] and [itex] x = -1 [/itex] because those values make the denominator zero. Since [itex] x^2 -1 [/itex] is not a common factor, the fraction cannot be reduced. The graph does not have a hole at any finite point (x,y). Values of [itex] x [/itex] that make the denominator zero cause the graph not to exist, even if there are no common factors in the fraction. [/QUOTE]
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How to determine a hole in a graph?
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