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

This is a video that depicts a problem that is not factorable like other concepts. This is a problem that involves imaginary along with real zeros. It takes everything that has been learned in the previous concepts and sandwiches it into one nice video. It explains everything step by step and helps you understand what is happening. It is a video about a problem with some imaginary friends that need to be found.

You have to pay attention to every single step that you are taking. If you mess up once then you are going to mess up the whole problem. Each step is a guideline for the next one. Be careful not to mix up the degrees or any of the positives and negatives. Do not forget to switch the signs when going from zeros to factors. It is very important that you take your time and really do each step with care and patience. It is a long amount of time for a single problem, but in the end it is really worth it. Be careful and enjoy.

SP#2: Unit E Concept 7 - Graphing a polynomial and identifying all key parts

Step by step: 1) Factor out the x.
2) Factor the rest of the equation.
3) List the end behavior (learned in concept 4)
4) List the x-intercepts (the zeroes) with their multiplicities (concept 6).
5) Find the y-intercept.
6) If possible find the extremas (the min and max) and then list in intervals of increase and decrease.
7) Plot all available points onto the graph, indicate your end behavior and then draw the graph following all graphing rules.

 

 

       This is an example of a factorable polynomial being graphed with all the appropriate parts. There are five steps that will always be there. Those are factoring the equation, end behavior, x-intercepts, y-intercept, and plotting/graphing them all. The parts that will sometimes be there are those of the extremas and the intervals of increase and decrease. This problem is to show you what these kind of problems look like and also what they consist of. They are multistep equations that involve your full attention. They are exciting problems that really review everything that we have been learning and introducing us to future concepts. They are great problems.

     This problem is not easy peasy though. You cannot breeze through these problems. You really have to pay attention to the end behaviors and the zeroes. The end behavior is based off of the highest degree and the leading coefficient in the standard equation. This one happens to be an even (the degree) positive (the leading coefficient). That means it will go up on both the left and right of the graph. Other equations will see different end behaviors though. Make sure your end behaviors are going in the correct direction and do not make up intercepts to keep in line with your end behaviors. The second is the zeroes. You have to make sure to put their multiplicities. That tells you how they will appear on the graph. If it is a single multiplicity then it will go straight through the x-axis, if it is a multiplicity of two then it will bounce on the x-axis, and if it is a multiplicity of three then it curves through the x-axis. These intercepts are the only times the graph will cross the x-axis. It will never go past a solid line with no intercept. Those are really the only tricky parts.

WPP#4: Unit E Concept 3 - Maximizing Area

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SP#1: Unit E Concept 1 - Graphing a quadratic and identifying all key parts

This is the whole problem. The work, the graph and the parent function along with the answers to create the graph.

This is the picture to show the steps and the graph to show you what the function becomes. You begin with your standard function. Then you add two to both sides of the equation. After that you take out the leading coefficient which is in this case -4. Make sure you have that on both sides of your equation to keep in line with the math rule, whatever you do to one side you must do to the other. Then you take your b. -3, and divide it by 2, giving you -3/2. Then you square your answer giving you 9/4. Add 9/4 to both sides within your parenthesis. Now you should have -4(x^2-3x+9/4)=2-4(9/4). After that you reduce your quadratic on the left and solve on the right. You should be left with the answer -4(x-3/2)^2=-7. Once you have that, you add 7 to both sides and get your parent function equation (also known as graphing equation) which is f(x)=-4(x-3/2)^2+7. From here you get your vertex (h,k) and that is (3/2,7) reduced is (1.5,7). Then you get your axis from that, which is the x-value so your x-intercept is 1.5. After that you go back to the equation right before the addition of the seven and find your x-intercepts by solving for x. You divide both sides by -4 then take the square root of both sides and finally add 3/2 to both sides. That makes your x-intercepts, with the 3/2 reduced, (1.5+√7/2,0) and (1.5-√7/2,0). To approximate that plug them both into your calculator and round to the nearest hundredths/thousandths place. Your x-intercepts become (2.82,0) and (.177,0). Then you find your y-intercept and to do that you plug zero into the x in the standard form and you get your answer of (0,-2). [The blue highlights are the parent function, the orange is your non-reduced and reduced x-intercepts, green is your vertex, pink dashes are your axis, and the purple is your y-intercept.] Then you plot each point, draw your dashed axis and connect the dots. Do not forget to reflect your y-intercept across the axis.

This is just the answers put into the key and color coded onto the graph for a nice visual. Your vertex in this equation is a maximum due to the negative a.

WPP#2: Unit A Concept 7 - Profit, Revenue, Cost

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WPP#1: Unit A Concept 6 - Linear Models

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