How to determine a hole in a graph?

  • #1
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0
I'm aware that in order to find the hole in a graph, you need to factor both the numerator and denominator, and look for terms that cancel out.

However, is it merely just looking for a term that cancels out, or is it more specifically a term that cancels out and makes the numerator equal to zero?

Thanks in advance.
 

Answers and Replies

  • #2
694
114
Can you clarify what you mean by a "hole" in a graph? If you mean points where the graph is not defined, then there are various scenarios where there can be non-continuity in the domain.
 
  • #3
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Can you clarify what you mean by a "hole" in a graph? If you mean points where the graph is not defined, then there are various scenarios where there can be non-continuity in the domain.
Something like this

upload_2015-1-18_13-13-9.png


Edit: Actually, this picture might have just answered my question.

So we have the equation: [itex]\frac{x^2-1} {x-1}[/itex]

which reduces to: [itex]\frac{(x+1)(x-1)} {(x-1)}[/itex]

The [itex](x-1)[/itex] terms will both cancel out, but that still leaves the numerator as a non-zero value. Thus, maybe it's merely just a term that cancels out that causes a hole. In this case, the term is [itex](x-1) = (x = 1)[/itex] which corresponds with the hole in the graph at [itex]x = 1[/itex]
 
Last edited:
  • #4
Stephen Tashi
Science Advisor
7,582
1,472
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.
 
  • #5
14
0
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.
Thank you, this helped!
 
  • #6
694
114
Adding to what Stephen said, explicit non-polynomial functions containing terms of the type ##u^{-|R|}## (where ##u## is an expression containing ##x##) have "holes" in their graphs if ##u=0## for any real x value . The simplest function of this type is ##y=x^{-1}## (u=x here) which has an asymptote at x=0. Similarly, by letting u=cos x and R=1 , the function will have multiple "holes" arranged in a recurring fashion wherever cos x = 0 (this will be the natural domain of sec x). I also must add that the concept is not just limited to fractions containing x terms in the denominator, but also logarithmic functions, where ##log_a u## is not defined for any x value where u=0.
 

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