How to determine a hole in a graph?

[hgducharme]

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.

 

[PWiz]

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.

 

[hgducharme]

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

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

So we have the equation: $\frac{x^2-1} {x-1}$

which reduces to: $\frac{(x+1)(x-1)} {(x-1)}$

The $(x-1)$ 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 $(x-1) = (x = 1)$ which corresponds with the hole in the graph at $x = 1$

 

[Stephen Tashi]

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 $f(x) =\frac{(x^2 + 1) (2x + 3)} { (x^2 + 1)}$ don't have a hole in their graph because (in the real number system) there is no value of $x$ 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 $g(x) = \frac{(x^2 -1 )(2x + 3)}{(x^2 - 1) }$ has a holes when $x = 1$ and $x = -1$.

The graph of the function $h(x) = \frac{ 6 + (x^2 )}{(x^2-1) }$ doesn't exist at the values $x = 1$ and $x = -1$ because those values make the denominator zero. Since $x^2 -1$ 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 $x$ that make the denominator zero cause the graph not to exist, even if there are no common factors in the fraction.

 

[hgducharme]

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 $f(x) =\frac{(x^2 + 1) (2x + 3)} { (x^2 + 1)}$ don't have a hole in their graph because (in the real number system) there is no value of $x$ 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 $g(x) = \frac{(x^2 -1 )(2x + 3)}{(x^2 - 1) }$ has a holes when $x = 1$ and $x = -1$.

The graph of the function $h(x) = \frac{ 6 + (x^2 )}{(x^2-1) }$ doesn't exist at the values $x = 1$ and $x = -1$ because those values make the denominator zero. Since $x^2 -1$ 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 $x$ 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!

 

[PWiz]

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.