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

Rational Functions - 2/5/13

We learned about rational functions. To understand what this phrase meant, we remembered that a rational number was simply a fraction with integer values in both the numerator and denominator.

Example of a rational number : 3 / 5

Rational fractions are very similar to this ration above, except the integers are replaced with polynomials in the numerator and denominator.

Example of a rational function: y =  x  /  (x+1)

If you take

y =  x  /  (x+1)

and graph it on your calculator, it will look like this (with zoom standard):



The graph shows two curves that are separated by a gap at roughly (-1 , 0). That means it’s a discontinuous function.

But we don’t always need the graph to find most of the characteristics of the function, which include:  x-intercept, y-intercept, horizontal asymptote, vertical asymptote.

I’ll use y= x / (x+1) to show how to find each. (this section can be found in our book starting on page 254)

X-intercept: Plug in zero for y. Then look at the numerator and see how you could get zero by plugging in a number for x. In this case, because the term in the numerator is only x, we plug in 0. So the x-intercept is at (0 , 0)


Y-intercept: Plug in zero for all the x terms. It would look like this : 

y = 0 / (0 + 1)

And that becomes y = 0, so the y-intercept is also at (0 , 0). By looking at the graph on your calculator, you can estimate the intercepts fairly easily, but with other function equations it’s more difficult to do so.


Horizontal Asymptote: Examine the “end behavior” of the function as it curves along the x-axis towards infinity (∞) and negative infinity. What y value on the coordinate plane is it seemingly approaching, yet never touching? If you look at the graph of y= x / (x+1), the function seems to move towards y = 1 in the negative infinity direction, and also rises towards (yet never reaches) y = 1 in the positive infinity direction. Therefore, the horizontal asymptote is :

y = 1

Another way to check this mathematically is to plug in a large number for all x-values in the function. Use a number like 1,000,000:

y = 1,000,000 / (1,000,000 + 1)    =    0.999999

This shows that the function rises towards 1 in the x direction for infinity.

Vertical Asymptote: The vertical asymptote occurs when the denominator equals zero. Plug in a number for x in the denominator to get zero:

y= x / (x+1)    plug in -1 for the x in the denominator

y = x / (-1 + 1)   

y = x / 0

This means that the vertical asymptote exists at x =  - 1

Here is a picture from pg. 264 in our book explaining the occurrence of asymptotes:

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Homework Problems
Our homework on Tuesday was pg. 268 #1-6, and 7-10. We tried some of these problems on Monday.

9).  f(x) =   -2  /  (x - 5)

X-intercept: 

0 =   -2  /  (x-5)

*because there is no x variable in the numerator, this function has no x-intercept

Y-intercept:

Plug in zero for all x-values

y = -2 / (0 - 5)

y = -2 / -5

So the y-intercept is at (0 , 2/5)

Horizontal Asymptote:

The function appears to be heading towards positive/negative infinity in the x-direction at y = 0

Plug in 1,000,000 for all x-values

y = -2 / (1,000,000 - 5)

y = -0.000002

That is very close to y = 0

Vertical Asymptote:

Try plugging in a number for x in the denominator so that it will equal zero:

f(x)  = -2 / (5 - 5)

f(x) = -2 / 0

Because 5 works to make the denominator 0, the vertical asymptote is at x = 5

Alright, I hope this post helped you understand rational functions.

 -Jesse