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Domain, Asymptotes, Intercepts of a Function

Date: 04/01/2003 at 22:33:31
From: Tom
Subject: Calculus

Hi, 

Consider the function: f(x)=x(x-12)(x+17)(x-8) divided by (x+4)

1) What is the domain of this function?
   What asymptotes does your function have?
   What are the x and y intercepts of your function?
2) Write this function in expanded form.
3) Find the first derivative of this function.
4) Find the critical values for your function.
5) Find the intervals where the function is increasing and decreasing.
   Find the relative maximum and minimum points of your function.
6) Find the second derivative of your function.
7) Find the intervals where your function is concave up and down.
   Find the inflection points of your function.


Date: 04/05/2003 at 19:37:29
From: Doctor Dotty
Subject: Re: Calculus

Hi Tom,

Thanks for the question!

The easiest way to check most of those things is using a graph.

      x(x - 12)(x + 17)(x - 8)
f(x)= ------------------------
               x + 4

When does the graph cross the x-axis? When f(x) = 0.

Well, the points where f(x) = 0 must occur when 

      x = 0, x = 12, x = -17, x = 8

It only crosses the y-axis when x = 0. This point has been covered 
above.

Now, are there any vertical asymptotes? Well, when x = -4 then the 
value of the function will be undefined. So there is an asymptote 
there.

Let's draw all that information on a graph. Where an axis is 
crossed, it will be marked with an "X"

                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
  -17          -4'    |      8         12 
  ---X-----------'----X-----X--------X-
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                
                 '    |                

As a fraction such as this tends to infinity, we can ignore every term 
except the one of the highest order on both the numerator and 
denominator.

To see why this is true, consider some examples:

   2 ^ 2 = 4
   2 ^ 3 = 8

The difference between these is pretty huge in relation to the value 
of 2^3. The difference is half of 2^3.

   10 ^ 2 = 100
   10 ^ 3 = 1,000

The difference between these is less huge in relation to the value 
of 10^3. The difference is a tenth of 10^3.

   1,000,000 ^ 2 = 1,000,000,000,000
   1,000,000 ^ 3 = 1,000,000,000,000,000,000

Here, the difference is much smaller in relation to the value of 
1,000,000^3. The difference is a millionth of 1,000,000^3.

As the number tends to infinity, this difference is so small that it 
is insignificant.

So:

      x(x - 12)(x + 17)(x - 8)
f(x)= ------------------------
               x + 4

The highest order of x at the numerator is x^4, and is x^1 on the 
denominator. Both do not have coefficients.

Therefore, as x tends to infinity:

      x ^ 4
f(x)= ----- = x ^ 3
        x

So, in the outer limits of the graph, it will become similar to the 
graph of x^3.

So the graph will enter the range of our axes from the bottom left and 
top right of the graph in the manner of a x^3 graph.

We will see what happens to the curve from right to left, as x is 
decreasing. So, looking from right to left, the curve enters from the 
top. 

We know it crosses the x-axis at x = 12, and again at x = 8. As there 
are no asymptotes between these, it must just duck below the axis at 
x = 12, and come back up again at x = 8. It then comes back down to 
cross both axes at (0,0), after which it is under the x-axis. 

It will now fly off downward as it approaches the asymptote at x = -4.

To determine what happens on the other side of the asymptote, we 
can follow the same sort of reasoning, but this time coming from
the left side of the graph.  We know that the curve starts below
the x-axis, and it crosses at only one place (x = -17), so it
has to fly off upward as it approaches the asymptote.


                .'    |                .
                 '    |                
               . '    |                
                 '    |                
             .   '    |                
                 '    |               . 
         .       '    |                
                 '    |   .             
  -17          -4'    | .    8         12 
  ---.-----------'----.-----.--------.-
                 '    |                
                 '  . |        .        
   .             '    |            .    
                 '    |                
                 ' .  |                
                 '    |                
                 '    |                
 .               '    |                
                 '.   |                

Can you check most of those points using this? You can check the 
first derivative by checking gradients, and the second derivative by 
checking when the graph is increasing and decreasing.

Does that help? 

If I can help any more with this problem or any other, please write 
back.

- Doctor Dotty, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Calculus
High School Functions

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