Rational Functions Continued

By Timmy Bollinger

Alright guys I am just going to pick up right where Jessie left off... In the last couple of days we have learned a couple of new things that have to do with rational equations: a short cut to finding the horizontal asymptote, and what a "hole" is and how to find its location and account for it on the graph.

The Horizontal Asymptote (H.A.) Short Cut:

               

                              Divide the leading coefficients in a rational function equation

      *This only works when the degree of the numerator is the same as the degree of the denominator*

   Ex.'s:

5x^2+3x-2

2x^2-7

For this example (^) the H.A.= 5/2 or 2.5

__x^2-9__

x^2-x-20

For this example (^) the H.A.= 1

3x^2-5

2x+1

For this example you need to put a place holder in to make the denominator's degree (x or x^1) match that of the numerator's (x^2). Adding the place holder makes the equation look like this:

__3x^2-5__

0x^2+2x+1

For this example (^) their is no H.A. because 3 / 0 does not work

However if the place holder (meaning the coefficient of 0) is in the numerator, then the H.A always = 0 because 0 divided by any other number is still 0

Identifying and Graphing "Holes" on Rational Functions:

A hole occurs when an identical factor appear on both the numerator and denominator of a graph

y= _2(x+2)(x-5)_

      (x+2)(x-3)(x-6)

As you can see the identical factor in this example is: (x+2)

This creates a hole on a graph because the equation places an intercept on an asymptote. When only the asymptotes and the intercepts are marked on the x and y axis, It looks like this:

In order to graph this equation, (x+2) must be cancelled from both the numerator and denominator.

The equation then becomes:

y= _2(x-5)_

      (x-3)(x-6)

*Even though your equation now works, you still have to mark the hole where the graph crosses a vertical line running through -2 on the x axis.*

In order to find out what the new y value of the hole is on your graph (it is no longer 0 because the x intercept: (-2,0) went away when you cancelled the factor (x+2)) you have to plug -2 in for x and solve for y. This looks like:

 y= _2([-2]-5)_

      ([-2]-3)([-2]-6)

y=-7/20

meaning the hole must be marked at (-2, -7/20)

Here is the function graphed with the hole marked correctly:

More information on Rational Functions can be found on pg 254 in the text books

-for specifically H.A. Short Cut: pg 262

-for specifically "holes": pg 267