WPP #6: Unit I Concept 3-5: Interests and Investments

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SV #4: Unit I Concept 2: Graphing Logarithmic Equations

     The viewer needs to pay attention on how to use the graphing calculator to give a visual of how the graph should be able to look like. When solving for x-intercept, don't forget to exponentiate! When solving for y-intercept, don't forget to PEMDAS!Also, you can either plug in "log" or "ln" in your graphing calculator because either way, they will both come out to be the same.***I forgot to mention the domain and range in this video as you can see. When you graph log equations, your range will have no restrictions so it will also be (negative infinity, infinity). It has no restrictions because your asymptote is x=h. For the domain, you will have to refer to your graph. In this case, the graph only goes from -5 to infinity (from the asymptote to infinity). The domain in this graph will have to be (-5, infinity). Thanks for watching the video!

SP #3: Unit I Concept 1: Graphing exponential functions and identifying x-intercepts, y-intercepts, asymptotes, domain, range

    The viewer must pay close attention when solving  for the x-intercept. If the log or natural log is negative, there is no x-intercept! Also, make sure to observe the problem before you actually start. There are so many small useful tips just by looking at the problem. You will eventually keep in mind of what the graph shall look like. Figure out if the graph will go above or below (negative "a" will go below). Find the asymptote so you can determine where or not there will be an x-intercept. From there, you can easily know your domain and range even before you start the problem!

Steps and Explanations!!
1. Label your a, h, b and k. In this type of equation, your asymptote will be y=k. 2. Find your x-intercept by plugging in zero for y. You will have to take the natural log in this problem in order to cancel and simplify your answer as much as possible.  3. Find your y-intercept by plugging in zero for x. Use pemdas to solve. 4. If an exponential graph has an asymptote that is y=k, then there will be no leading restrictions on the domain. Thus, it would be negative infinity, infinity (in order pairs). 5. Determine the range simply by looking at your graph. In this case, the graph is above the asymptote. So you will have to write your range practically based on your observation of the graph. The graph began at 3 and up beyond to infinity (3, infinity).

SV #3: Unit H Concept 7: Finding Logs with Given Approximations

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 In this concept, finding logs with given approximations provide the clues in order to be able to evaluate the logs in the problem. The "clues" will guide you in terms of knowing what values or letters will equal to a certain log. Using the properties of logs from the previous concept, this problem will show you how to find your answer just based on the clues that are given. This concept combines all the other previous concepts in this unit.
 The viewer needs to pay special attention to special hints. Hints such as log base b of b equal to one and log base b of one equal to zero. Make sure to break down the problem and see what kind of factors go into the numerator and the denominator of the problem. Also, if you have log base b of b to the power of x, make sure to bring the exponent down and in front of the log! This is an example of the power law. Good luck and thanks for watching the video!

SV #2: Unit G Concept #1-7: Rational Functions

This problem is everything about rational functions! You will learn how to find your asymptotes, holes, domain, x-intercept, and y-intercept. Each asymptote, such as the horizontal, slant, vertical, and hole, has unique rules and steps to determine the equation. Most of the different asymptotes depend on the degree of the top and bottom of the polynomial. This concept also includes the limit notation.

The viewer needs to pay special attention the vertical asymptote and hole(s) because for vertical asymptote, the denominator (or bottom) has to be set up to 0. The tricky part is the denominator. Why? Because the viewer needs to make sure that whatever is actually left in the denominator, meaning everything else is already checked if it can be canceled, can be equaled to 0. Also, it is relatively easier if you use the simplified equation for the y-intercept since there will be a hole in this problem.