Arlene S.'s WPP#6: Concepts 3, 4, &5

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Student Video #3: Unit H. Concept 7

• What is this video about?
• This video goes over how to find logs with given approximations/clues. 
• What does the viewer need to pay special attention to in order to understand the concept?
• Remember the properties of expanding logs. [* Product Law: you break the logs apart and add them separately. *Quotient Law: You separate the terms of the log by subtracting them. * Power Law: you bring the power that is raised to the front of the log.] Also, note that if the base is the same as the answer, the exponent is 1 ---> logbb=1

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Unit G Summary Question #10: Rang of a Rational Function.....

10. While the domain of a rational function depends on DIVAH, what do you think the range of a rational function depends on. Give an example.

NOTE: D-omain
               I-s the
              V-ertical
              A-asymptote & the
              H-ole

I believe that range of a rational function is dependent upon all of the y-values that belong to the x-values that come from the vertical asymptotes and holes. For example, the vertical asyptotes for a graph were to be x=-2 and x=2 and the hole would be x=1. This would make the  the domain, or the bad values, -2,2, and 1. With that, the range would be the infinite numbers that these numbers passed from +infinity (the + numbers of the y-axis) and - infinity (the - numbers of the y-axis). Because the lines run vertically downwards, the coordinate pairs would be something like (-2,4), (-2,3), (-2,2) ,(-2,1) ,(-2,0) ,(-2,-1) ,(-2,-2), (-2,-3) and so on upwards and down the y-axis, with the domain staying the same. This would be the same for the x=2 vertical asymptote. In short, the range would be ALL REAL NUMBERS. With that, the range of a hole would simply be the y-value. For exapmle, the hole would have the cordinates of (1, -1/2), -1/2 would be the range.

As you can see in this image, the vertical asymptote is
x=-1. The range, or y-values, run up and down make it
ALL REAL NUMBERS.

On the other hand, if the y-value was a horizontal asymptote, the line would run straight across the y-axis and so it would NOT be all real numbers. For example, if y=2, the y-value would remain constant whereas you x-value would be constantly changing as you traveled across the graph from left to right. [(1,2),(2,2),(3,2),(4,2),(5,2), etc.,...] thus, the range of a horizontal asymptote would simply be the value of y. We have already been taking notice of this and it is noted in our limit notation. (If needed: Review the lesson in Unit G Summary Question #2).

Unit G SUmmary Question # 9: X-Intercepts of a Rational Function

9. Describe how to find the x-intercepts of a rational function. Include both the long way and the shortcut way, explaining why the shortcut makes mathematical sense.

How to: The LOOOOONG WAY....
In order to find the x-intercepts through the use of the faster method, it is important that you know how we got to that point and why.
To make it easier, lets use an example: 3x+3/x^2-4
     ^* REMEBER TO FACTOR!--->  3(x+1)/(x+2)(x-2)
       * now we get to the fun part..... x-intercepts!!! :D
       First: set the y [or f(x) ] to zero, thus setting the equation to zero
                                    ^   0= 3(x+1)/(x+2)(x-2)
     
       Second: Mulitply both sides by the denominator
                                  ^   (x+2)(x-2) 0=3(x+1)/(x+2)(x-2)    x    (x+2)(x-2)
     
       Third: As you can see, the denominator cancels out on the right side
                                 ^ (x+2)(x-2) 0=3(x+1)/(x+2)(x-2)    x    (x+2)(x-2)

                                               * now you are left with ....
                                                   (x+2)(x-2) 0=3(x+1)
       
         Fourth: as you can see, the (x+2)(x-2) will be eliminate becauseANYTHING                 multiplied by zeor will turn to...... zero!

         Fifth: Now you have:
                                                          0=3(x+1)
     
          Sixth: divide by 3:
                                                          0=3(x+1)
                                                          3       3

       Seventh: solve like you normally would.... in this case subtract 1 to both sides
                                                          0-1=x+1 -1
 
      Eighth!!! And your x-intercept is....      x=-1  making it (-1,0)


How to: SHORT WAY...
Phew! Now that you know how to do it the long way, the shorcut will make a lot of sense! :)

          First: set the y [or f(x) ] to zero, thus setting the equation to zero

                                             ^   0= 3(x+1)/(x+2)(x-2)

         Second: since you know that the denominator will always cancel and be eliminated you can keep on moving forward. SIGH OF RELIEF....

                                                      0= 3(x+1)/(x+2)(x-2)

         Third: You are left with...

                                                       0=3(x+1)

       

          Fourth: divide by 3:

                                                          0=3(x+1)

                                                          3       3

      Fifth: solve like you normally would.... in this case subtract 1 to both sides

                                                          0-1=x+1 -1

   

      Sixth!!! And your x-intercept is....      x=-1  making it (-1,0)

       

Unit G Summary Question #6: Plotting Holes

6. How do we find the appropriate place to plot a hole if the y-value is undefined when plugged into the original equation?

NOTE: If you need to review how to find a hole, refer to Unit G Summary Question #4 :)
            * Hopefully you noted that you only have a hole if the something that was the same on the top    and bottom was canceled.
            * Holes are in place of a regular point due to the fact that the denominator (value on the bottom of the fraction) cannot be equal to zero.

Notice how the hole is like a break in the graph, but the
graph continues after skipping the spot where the hole
was plotted. As a result, the whole is NOT included
within your answer. It also acts as a bad value thus it is
annotated in your domain of a graph.
After finding the x-value of the hole from the original equation and you plug it back into the original equation, you will get the answer as being undefined. In order to find the y-value CORRECTLY, you will have to plug that x-value into wherever the 'x' shows up in your simplified equation. This will give you your y-value and thus your coordinates of the hole. When finding the y-value of the hole, if it is undefined then that basically means that the hole is NOT included within your answer. When plotting the hole, make sure that you do so with an OPEN circle -∘- (not a just a regular point ⋅).