pre-calculus graph

Graph of a rational function can be discontinuities because it has polynomial in the denominator. Is possible value x divide by zero?

Example, there is a function f, it is f of x equals x plus two divided by x minus one. If we insert x equals one, we get the value of f of one equals one plus two divided by one minus one equals three divided by zero. We know that three divided by zero is bad idea and the graph f of x equals x plus two divided by x minus one will break in function graph. If we insert x equals zero, we will get the function f of zero equals zero plus two divided by zero minus one equals negative two.

We will draw graph f of x equals x plus two divided by x minus one in the x y-coordinate plane. First we draw x y-coordinate plane, then we draw the graph f of x equals x plus two divided by x minus one as we know that this graph intersects axis y at point (0, -2), then we can draw the graph which through point (0, -2) and approach axis x in axis positive y. So we can say that graph f of x equals x plus two divided by x minus one is break. On the contrary, if there is line x equals one, so the graph approaches this line and axis positive x is discontinue.

Rational function does not always work this way!

Not all rational functions will give zero in denominator. Take the graph f of x equals one over x square plus one, it’s denominator never zero because of plus one and of course this graph is not break in function graph.

Do not forget!

Rational function denominator can be zero if the polynomial have smooth and unbroken curve and for rational function x equals zero in denominator, because that is impossible situation. It is impossible because there is not value for the function, so make break in function graph.

The break is showed up by two ways!

First, it can break in graph function because of the missing point in the graph. For example the graph f of x equals x square minus x minus six over x minus three has missing point at line x equals three. It can be happened because if we insert x equals three to the graph, so we get f of three equals three square minus three minus three over three minus three equals zero over zero. It can break because the value zero over zero is not possible, not feasible and not allowed. Typical f of three equals three squares minus three minus three over three minus three equals zero over zero named missing point syndrome. If the result is zero over zero, to solve this problem we can make the factor top and button to be simplify. For example, y equals x square minus x minus six over x minus three. We get the top factor are x minus three and x plus two, the button factor is x minus three. So we can write y equals (x minus three) times (x plus two) over x minus three, and we can divide numerator and nominator with x minus three so we get y equals x plus two. Now, it is not problem if we insert x equals three to the new function y equals x plus two.