CLICK HERE!
In this video, I will be going over the process of using matrices and the Gaussian Elimination process in order to find the correct values for the consistent independent ordered pairs. There are my tricky things in this video that we must pay attention to, so make sure that you pay attention to everything in general. To be specific though, remember that there are four steps you need to take in order to find the correct functions, which means that you must remember them in order to complete the matrix. Along with that, remember how to check your answers with reduced row-echelon form to save you time in case you make a mistake along the way. Yes, fractions are back, but that does not mean that you will get decimals in your values. In fact, make sure you only get whole numbers as we are only working with whole numbers for now. Other than that, thank you for watching! :)
Showing posts with label SV. Show all posts
Showing posts with label SV. Show all posts
Monday, November 11, 2013
Saturday, October 26, 2013
SV #4: Unit I Concept 2: Graphing Logarithmic Functions
Click Here
This video goes over logarithmic functions and also includes the parts that help find the graph to it. There are many key factors that you must pay attention to because this time around there's a log involved. Given this information, you know that for the x-intercept, we have to exponentiate in order to get rid of the log. Also, for the y-intercept, you must remember to change the base in order to figure our the ordered pairs. Also, in order to graph the equation, you need key points of significance. However, in order to get thoe points, you first need to plug it into your calculator the way I described in the video, so please make sure to do this before continuing. Lastly, as a clarification, the graphing calculator will not entirely graph the equation correctly, so the part left of -5, which is the asymptote, will be blank because of that same reason. Don't worry though, you do not need to graph that section. Other than that, thank you for watching :D
This video goes over logarithmic functions and also includes the parts that help find the graph to it. There are many key factors that you must pay attention to because this time around there's a log involved. Given this information, you know that for the x-intercept, we have to exponentiate in order to get rid of the log. Also, for the y-intercept, you must remember to change the base in order to figure our the ordered pairs. Also, in order to graph the equation, you need key points of significance. However, in order to get thoe points, you first need to plug it into your calculator the way I described in the video, so please make sure to do this before continuing. Lastly, as a clarification, the graphing calculator will not entirely graph the equation correctly, so the part left of -5, which is the asymptote, will be blank because of that same reason. Don't worry though, you do not need to graph that section. Other than that, thank you for watching :D
Thursday, October 17, 2013
SV#3: Unit H Concept 7- Finding logs given approximations
CLICK HERE
This is problem goes over how to find logs based on approximations. Before you start remember to add logbB=1 and logb1=0 to your given clues in case you need them. For this particular problem, we will only be using the property that equals 1. Another important thing that you must remember is the difference between product and quotient logs. Also, just to make this problem easier, divide the log your are looking for by the given clues so you won't have to take much time trying to figure out which factors will work for the problem Logs that have addition in between them are product (vice versa) and those that have a subtraction sign are quotient (vice versa). When your expanding, you should have one log for each number so you can then substitute in the values or letters that were given at first. I did not mention this is in the video, but implied that each number has to have a log, so I hope that makes sense now,if it didn't before. Thank you for watching!
Again, I am sorry for the mistake. Basically the only difference is that all the factors on the denominator will have subtraction signs as it is a quotient considering that is underneath the numerator.
Sunday, October 6, 2013
SV#2: Unit G Concepts 1-7- Finding All Parts and Graphing a Rational Function
SV #2: Unit G Concept: 1-7 Rational Functions
Click HERE to watch :)
This video goes over rational functions and as well includes the slant asymptote, vertical asymptote, holes, domain, intervals, and explains how to graph it. This functions does not include a horizontal asymptote because its leading degree on the numerator is bigger than the leading degree on the denominator. Therefore, this graph has a slant asymptote with one hole and two x-intercepts. Along with that, this video goes over notations, equations, and ordered pairs.
In order to understand this problem better, you must note that the slant asymptote does not require you to continue solving past the needed y=mx+b equation. Along with that, because you have the same factors on the numerator as well as the denominator, the graph is going to have a hole. The graph will then skip the ordered pair as it is a hole. Lastly, it is very important to remember that the graph does not touch the asymptote even though is might seem as it does.
****Correction:
I'm sorry for the mistakes, but here is a correct version of the graph. For the holes, the canceled factor, which leads to the equation x=-3, is the x-intercept for the ordered pair. You just have to plug in the x-value in the equation as shown in order to get the y-value of -3/4. This means that the hole on the graph changes, but you still have to skip it as it does not exist on the graph when you trace it (please see the graph from the picture). As for the y-intercept, you have simply plug in 0 for the x-values in order to find the ordered pair of (0,0). Other than that, the rest is correct.
Sunday, September 29, 2013
SV#1: Unit F Concept 10- Finding all zeroes to a 4th degree polynomial
Click HERE to watch video :D
This video covers how to find all zeroes to a polynomial of 4th degree. It will start by finding all the possibilities of the coefficient and the leading degree. Of course, there were a lot of possibilities, so we narrowed down the possibilities by using Descartes Rule of Signs. That narrowed down our possibilities and we then went straight into trial-and-error with the different zeroes using synthetic division. After bringing down the equation to a quadratic equation, we used the quadratic formula to find the remaining zeroes. We carefully solved the equation and set them x-(...) to find the exact zeroes.
The viewer has to note that the odd degree changes signs, but that the even degree does not, just as it was mentioned in the video. Along with that, one must acknowledge that not all the possibilities work and that a only a process of trial- and-error will help lead to the quadratic equation. Also, remember that the the radical cannot be left as a negative, so there will be imaginary numbers. This goes back to the reason why we tend to count down by 2's during Descartes Rule of Sign. Lastly, the factors can be written as a quadratic equation, but it can also be left as the answer one gets after completing the quadratic formula.
*Here's a correction, sorry about the mistake.
Student Video #1
This video covers how to find all zeroes to a polynomial of 4th degree. It will start by finding all the possibilities of the coefficient and the leading degree. Of course, there were a lot of possibilities, so we narrowed down the possibilities by using Descartes Rule of Signs. That narrowed down our possibilities and we then went straight into trial-and-error with the different zeroes using synthetic division. After bringing down the equation to a quadratic equation, we used the quadratic formula to find the remaining zeroes. We carefully solved the equation and set them x-(...) to find the exact zeroes.
The viewer has to note that the odd degree changes signs, but that the even degree does not, just as it was mentioned in the video. Along with that, one must acknowledge that not all the possibilities work and that a only a process of trial- and-error will help lead to the quadratic equation. Also, remember that the the radical cannot be left as a negative, so there will be imaginary numbers. This goes back to the reason why we tend to count down by 2's during Descartes Rule of Sign. Lastly, the factors can be written as a quadratic equation, but it can also be left as the answer one gets after completing the quadratic formula.
*Here's a correction, sorry about the mistake.
Subscribe to:
Posts (Atom)