How to determine a hole in a graph?

The key is to find the values of x where the function is undefined, not just where the denominator is zero. In summary, when looking for holes in a graph of a function that is the ratio of polynomial factors, you must consider all values of x that make the denominator zero, as well as any other values that would make the function undefined. This includes values where the numerator becomes zero, or where the function contains terms that are not defined for certain values of x.
  • #1
hgducharme
14
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.
 
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  • #2
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
PWiz said:
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
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
Stephen Tashi said:
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
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.
 

1. How do I identify a hole in a graph?

To identify a hole in a graph, you can look for a point where the graph appears to be discontinuous or has a break in its continuity. This can be visually seen as a gap or an empty space in the graph.

2. What causes a hole in a graph?

A hole in a graph is caused by a point where the function is undefined or does not exist. This can happen when the denominator of a rational function is equal to zero, resulting in a vertical asymptote.

3. How can I determine the coordinates of a hole in a graph?

To determine the coordinates of a hole in a graph, you can set the denominator of the function equal to zero and solve for the variable. The resulting values will be the x-coordinate of the hole. Then, substitute this value into the original function to find the corresponding y-coordinate.

4. Can a hole in a graph ever be filled?

No, a hole in a graph cannot be filled. It represents a point where the function does not exist and cannot be evaluated. However, you can approach the hole from either side and the function will have a limit at that point.

5. How is a hole different from a removable discontinuity?

A hole is a point where the function is undefined, while a removable discontinuity is a point where the graph is undefined but can be "filled in" by defining the function at that point. In other words, a removable discontinuity can be removed by redefining the function at that point, while a hole cannot be filled.

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