SV #5: Unit J Concept 3-4 - Solving 3-variable systems

     The viewer needs to pay special attention on the order and patter we are solving to get the 3 0's. We start with Row 3 Term 1, then Row 2 Term 1, and finally Row 3 Term 2. After the 3 0's are solved for, we continue solving by making a 1 stair-step pattern. We do this by multiplying the reciprocal of the leading coefficient, or by dividing by the leading coefficient. Probably the most important thing overall is that this is solving to get consistent independent solutions (one answer). Consistent independents solutions are a bit different than solving for consistent dependent solutions (infinite solutions). 

SV #4: Unit I Concept 2 - Graphing logarithmic equations

   The viewers need to pay special attention on how the asymptote is found. The asymptote equation is x=h and h comes from the "base of" portion of the log and is equal to 0. In this video for example, the "base of" is x+2, it is equal to 0 and solve algebraically to get the asymptote of x=-2. The viewer also need to pay attention to how the x and y intercepts are solve. For the x intercept, the equation is equaled to 0 and solved algebraically, however exponentiate is required. For the y intercept, x is 0 and solved algebraically; in this video log base b of b=1 property is used. And finally, the range is always (-infinity, infinity), but the domain is (asymptote, infinity).

SV #3: Unit H Concept 7 - Finding logs with given approximations

          The viewer needs to pay special attention to the approximations that he/she would use and the ones on the video. There are several ways to factor the logs, and the approximations, itself, can be factored (for ex. in this video: 4 and 2). Remember that when expanding logs: multiplication corresponds to addition; division corresponds with subtraction; and exponents move to the front of the log and act like coefficients. And finally, one must know and remember that logbb=1 and logb1=0, which serves as two extra approximations.

SV #2: Unit G Concepts 1-7 - Finding all parts and graphing a rational function

     This problem is about graphing rational functions. The equations is a fraction with polynomials on both numerator and denominator. The problem contains several steps in order to graph the function and it's asymptotes. In order to graph it, we must find the domain, x-intercept(s), y-intercept(s), and a few points. This problem also contains:

• a numerator with a degree of 3
• a denominator with a degree of 2
• one hole
• one vertical asymptote

The viewer needs to pay special attention to the degrees to understand the rules of finding the asymptotes.(There is no horizaontal asymptote because the degree is bigger on the top. To find the slant asymptote,, long division is used. Factor and simplify the equation, and make the denominator equal to 0 to find the vertical asymptote. The common factors are equaled to 0 and are the holes.) The viewer also needs to pay attention the graph. In order for the graph to fit, the intervals were changed: x by 1s and y by 2s. Remember that DIVAH stands for Domain Is Vertical Asymptote(s) and Holes.To find the x-intercept(s), y equals to 0, and vice versa.

SV #1: Unit 7 Concept 10 - Finding all real and imaginary zeroes of a polynomial

      This problem is about solving for zeroes and factorizing the polynomial: -35x^4+89x^3-110x^2+63x-7. The zeroes for this type of polynomial will result to be real and imaginary/complex. The factorization will also result as complex for the most part. This problem will have multiple steps starting with finding the possible real/rational zeroes and finding the possible positive/negative real zeroes. From there, the possibilities make a more precise attempts to find the zeroes, and lead to factorization.

     The viewer needs to pay special attention in finding the possible zeroes: real/rational and positive/negative. Finding the real/ration zeroes involves p/q, which is the factored last term over the factored first term. Find the possible positive and negative zeroes involves Descartes' rule of signs. The positive and negative possibilities have a small different rule throughout the process. The last thing views should pay attention is that the quadratic formula is easier to use when the polynomial is reduced to the 2nd degree and there can be no negative in the radical.