Jorge's Class
Daily Updated Notes
Started Off...To get ready for the test, we were given three equations and having to use the methods that we've learned so far to solve them. Graphing them, finding x & y intercepts, minimum/maximum, and vertex. These type of problems will be on the test and so by doing this we get extra practice. NotesNo notes for today! Remember that if you go to dance tonight you ear 15 extra credit points for attending!
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Started Off...Continued with the quadratic equation page and if you were finished you could work on honors for math. NotesNo notes, just that the quadratic formula page is due at the end of class Friday!
Started Off...We worked on the worksheet that was passed out yesterday during the whole period, asking questions when we needed and asked each other for help. NotesNothing new was taught until #15 and over, there were more steps to complete with the equations given to you.
Started Of...Started off in class with reviewing the worksheet that was given out yesterday. Answered questions that students had and moved on to a new concept. Learning how to go from standard from to vertex form and how to factor it out and find the x intercepts. There was also another worksheet passed out, that is not due as far as we know. NotesSame concept as yesterdays, but with negatives.
Also, and equation (b/2)^2 is helpful. x^2 + 2x - 3 = 0. First add on both side to get x isolated. But. when you do that, another number will replace where 3 was. You'll replace it with 1. Now you have a oerfect square, meaning you can put it into standard form. x^2 + 2x + 1 = 4 ---> (x+1)^2 = 4. Now square root both sides to get rid of the squared, and you will have a negative & a positive 2 = x+1. Subtract 1 on both side. 2-1 = 1 & -2 - 1 =-3. -3 and 1 are your x intercepts. Started Off...Started off in class with a warm up, eventually leading up to the lesson that we learned about today. The lesson was on finding x intercepts/roots/zeros/solutions, how standard from equals to vertex form and equaling them to zero. Also, there was a worksheet that was passed out and was only classwork. Notes Standard Form - x^2 + 8x + 15 Vertex Form - (x+4)^2 - 1 Are these the same? Are the equal to each other? If so how? Why? We answered tehse questions. Starting with standard form, you can factor that out easily. (x+3)(x+5) and equal them to zero. Giving you x = -3 & x = -5. Lets see what we get with the vertex from. First you want to get rid of the -1 that is there because it'll get in the way of the next step. So you add +1 on both side. Now you square root both side to get rid of the squared. All you have now is 1=x + 4. BUT the 1 is now a negative AND a positive (because a + & + is a + and - & - is also a +) which is crucial. Subtract 4 on both sides meaning 1 - 4 = -3 & -1 - 4 = -5. -5 & -3 are your x intercepts which is what we got with the standard form. Started Off...Started off in class with a warm up and went over it. For the rest of class we self graded our tests from last Thursday. NotesThe photos of the warm up are just explaining that there are two answers for each of the problems.
Started Off...Started off class with going over what we learned the day before. How to find all four factors of graphing quadratics (x intercept, y intercept, vertex, up/down). We then continued with the same worksheet until the end of class NotesThe worksheet that was passed out yesterday will be due on Monday #'s 1-20. The photo below is just a quick review on how to find all four factors of parabolas.
Started Off...Started off class with another warm up, focusing on square roots to get better practice with them. We then went over with what learned about how to graph quadratics at a faster pace with doing little work. A worksheet was then passed out as a practice sheet, which is still unknown when it'll be due. Notesx intercept - factor out the equation, making each one equal to zero y intercept - will always be the last number on the equation (or the c) up or down - if the first number/constant is positive, the parabola will be in a smiley form unless it is negative and it'll be in a frown vertex - (-b)/2a this the equation to find out out the vertex of your quadratic equation. ex; x^2+3x-4 ---> b = 3 ---> -3/2(1) = -1.5 *this is your x coordinate of your vertex (-1.5)^2+3(-1.5)-4 ---> -2.25-4.5-4 = -10.75 *your y coordinate of your vertex vertex =(-1.5,-10.75) Started Of...Started off with a warm up about factoring. We then went over about parabolas again, but figured out easier ways to graph the equations without doing so much work. Afterwards, we were assigned 4 equations to solve the x intercept, y intercept and the vertex of each one. NotesTo find x intercept - factor out the equation, then make them equal to zero to find both coordinates
ex; y=x^2+2x-3 ---> y=(x+3)(x-1) x+3=0 / x-1=0 ---> x= -3 & x =1 ---> (-3,0) (1,0) To find y intercept - will be the last number on the equation (b) ex; y=x^2+2x-3 ---> is your intercept To find the vertex - will be the median between the two x cordinates that you have ex; -3 &1 ---> -3,-2,-1,0,1 ---> is your vertex on the x axis using 'common sense' depending where the lowest coordinates are, you can figure out what the y part of the coordinate is (example below) |
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December 2016
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