Question 597013
A "hole" occurs in a rational function when the numerator and denominator both have a factor that could be zero. The "hole" occurs at x values that make these factors zero.<br>
For example:
{{{y = (2x^2 + 2x)/(5x+5)}}}
Factoring we get:
{{{y = (2x(x + 1))/(5(x+1))}}}
The numerator both have a factor of (x+1) and (x+1) could be zero so there will be a hole in the graph. The hole will occur at the value(s) of x that make (x+1) zero. That would be x = -1. So the hole in this graph will be at x = -1.<br>
Remembering not to forget the hole, we can now cancel the common factor leaving:
{{{y = (2x)/5}}}
or
{{{y = (2/5)x}}}
This is the equation of a line with slope of 2/5 and a y-intercept ... <i>and a hole at x = -1!</i><br>
On the other hand,
{{{y = (2x^3 + 2x)/(5x^2+5)}}}
factors into
{{{y = (2x(x^2 + 1))/(5(x^2+1))}}}
Again we have a common factor. But this time <i>it cannot be zero!</i>. So there will not be a hole. We can still reduce the fraction:
{{{y = (2/5)x}}}
The graph of this is the same line as above except that there is no hole.<br>
To manufacture a rational function with a given hole, just start with a numerator and denominator that will be zero at the given value: (x - holenumber), For a hole at x = -57 we use (x - (-57)) or just (x + 57):
{{{y = (x+57)/(x+57)}}}
and then multiply the numerator and denominator <i>by different expressions</i>. These different expressions can be whatever you like (except a zero). Since you should multiply out the numerator and denominator I would recommend fairly simple expressions. Now we have:
{{{y = ((expression1)*(x+57))/((expression2)*(x+57))}}}
where "expression1" and "expression2" are the two different expressions you've chosen. Multiply out the the numerator and denominator and you're finished.<br>
If you want multiple holes, just create multiple factors. For example if you want holes at -57 and 7 you would use:
{{{y = ((expression1)*(x+57)*(x-7))/((expression2)*(x+57)*(x-7))}}}