Monday, September 30, 2013

Student video #1 Unit F Concept 10 4th and 5th degree polynomials

 For this student problem we're focusing on finding real and complex zeros of 4th degree polynomials. To solve we must combine our knowledge from concepts in this Unit together all in one problem. This video should show you the appropriate steps to reach the zeros.

            The viewer needs to pay special attention to the very end in which we multiply x by our denominator.  Viewers are also to pay close attention the synthetic division to make sure we find our zero heros. You must also pay close attention to when the equation reaches a quadratic so we can factor or plug it into the quadratic equation

Monday, September 16, 2013

SP#2: Unit E Concept 7: Graphing polynomials, including x-int y-int, zeros (with multiplicities), end behavior. All polynomials will be factorable.

The work shown on the right are the steps to solve. First you must take the multiplicities and convert them into factors. 4 becomes x-4 repeating once. With that, we multiply two separate equations times each other to get our equation. The factors are also the x intercepts. To get the y-intercept we plug zero into all the x values and get what's leftover. In this case 32 is our y intercept. Our factored equations are our factors of the multiplicities.

This problem is about finding the equation, factored equation, end behavior, x intercepts and the y intercept. This problem is also based on knowing how to factor correctly and graphing the equation.We must also know the leading coefficient and degree in the equation.

Things to take note of are the leading coefficient and degree in the main equation.  Another thing to be wary of how the end behavior works. One key thing to pay close attention too is that the factor of the multiplicity is correct.

Monday, September 9, 2013

WPP #3 Unit E Concept 2 identifying maximum and minimum values of a quadratic application

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SP#1: Unit E Concept 1: Identifying x-intercepts, vertex, axis of quadratics and graphing them. Quadratrics in standard form

1. First step to solve is to add a 12 so the equation becomes -6x^2+24x=12
2. Then factor out a -6 while putting a -6 on the other side. -6(x^2-4x___)=12 -6(__)
3. divide the -4 by 2 then square it which becomes -4/2=-2^2=4
4. Plug the four into both sides and then simplify.
5. The equation then becomes -6(x-2)^2=-12
6. Divide -6 so the equation becomes (x-2)^2=2
7. Square root both sides which becomes x-2= square root of 2
8. Once the square root has been distributed the answer becomes + or - square root of 2.
9. Add 2 to the other side.
10. The equation is now x= 2+ square root of 2
and x= 2- square root of 2. These are the x intercepts.

To find the parent equation you take the equation -6(x-2)^2=-12 and add 12
To find the vertex you use the form (h,k) which is 2,12
To find the y intercept you plug in 0 for all x values of the standard equation.
To find the axis, the value is h of h,k which is 2

  The problem is about finding the parent graph equation. Also other things that must be found are the vertex, and whether it is max or min. The y-intercept must be found along with the axis and the x intercepts.

The viewer needs to pay close attention to the steps of whats being factored and the steps taken afterwards. Important things to take note is the b/2^2 part. Another thing to pay close attention is when you simplify you can simply take b/2 and use that as the equation.