Point Slope 2

Today in math class we went over point slopes and linear graphs again. We learned that a linear graph will always be a straight line, a absolute value graph will always be in a "v" shape, and that a quadratic graph will always be in a "u" shape. Whenever we see y = x, then it is a linear graph. Whenever we see y = x² , it is a quadratic graph. When ever we see y = IxI, it is a absolute value graph. Then after all of that was settled, we went on to just goin over what we did yesterday. We then graphed and did pretty much what we did yesterday.

Point Slope

Today in class we learned about point slope. The formula for point slope is:  y – y¹= m (x - x¹). So If I gave you something like this:  3/2;(4,8), you would substitute the values into the equation. y - 8 = 3/2x - 12/2 ----> y = 3/2x -6 +8 ----> y = 3/2x +2. So point slope isn't as bad as it sounds. All you mainly have to do is memorize the formula and substitute the values in. Since 4 in (4,8) is x, in the equation you would substitute 4 in x¹. Same thing goes for  8. You would 8 in y¹. Also your fraction would be your slope and in the equation y = mx + b, you would substitute m as 3/2.

Slopes 4

Today in math class we continued on slopes. Mr. S was gone for three days because he was really sick so almost a week without notes. He gave us some problems and we had to find with point doesn't belong by graphing and finding the slopes. To give you an idea, here's an example:

It's really all we got to today because some students in my class were confused. Then there was a fire drill that wasted 10 minutes of our learning time. Not really the best day ever.

Slopes 3

So today in class we continued to work on slopes. Mr. S gave us two input output tables with some missing numbers in the "y" column and we had to find the rule. The first rule was y = 4x-3. The second rule was y = 3x-2. Then we had to find the slope and graph it. So all we had to do was follow a formula and voila! We had our slope and we graphed it. So here is a visual picture of what I just explained for all you visual learners:

Then after that he gave us a mini-poster project due tomorrow. Pretty easy if you ask me.

Slopes 2

Today in math class we continued on learning about slopes. Once we got into the classroom he gave us a Do Now, which is just a problem we had to solve once we get into the classroom. The problems were pretty basic and easy to do. The first one was y = 3x-3. It was a blank input output table. We filled the first column, "x", with numbers from 1-6. Then we took the equation on top of the chart and filled the second column, "y", with 0,3 ,6, 9, 12, and 15. Then the second problem was y = 2/5x+2. We filled the "x" column with 1-6 like the first one. For the "y" column we put 2 2/5, 2 4/5, 3 1/5, 3 3/5, 4, and 4 2/5. Again, I know this is the worst explaination ever so here's a picture!

Slopes

So today in algebra we started a new unit. First he gave us an input output table with some missing spots in the "y" column. We then had to solve for "y" and then plot the points on a graph. Then when you connect the dots it should make a slope. A slope is really just a straight line. Mr. S then gave us a Co-Planar. Its basically just the steps we go through.
1) Locate y-intercept. 2) Use slope. 3) Locate three points. 4) Connect the dots. Our problem was 3/5x -3. So you move three points down. Then you move five points to the right and left. Then from the spot after coordinates (0,-3) move three points down. Your new coordinate will be (0,-6). After that look for (0,-6) and (5,0) and make a mark where they intercept. When you conncet points (-5,0), (0,-3), and (5,-6) it will make a slop. Okay I know that sounds really confusing maybe but teaching is hard! I don't know how Mr. S can do it all day! So here you go:

So here's how it SHOULD look like. FYI: if a line points left (down), its a negative line. And then if iit points right (up), its a positive line.

Function Notation & Matrices

Today in class we were told to find "f(x)". Since "f(x)" is just another way to say "y", all we had to do was substitute "x" with three since it said: Find f(3). Then we had to simplify all the equations and then multiply after that and then find the determinant. As always, I'm bad at explaining things so here is the problem in all of it's glory:

So after countless and countless attempts to find out this answer, it turned out that Mr. S had to give me the answer. I really hope I'm ready for the test tomorrow. Well, ready for it or not, bring it on!

Review!

So today in class we mainly reviewed for our math quiz tomorrow. We also did more exercises on multiplying matrices with negatives. And guess what? I think I've got all the negatives down! Finally! So we also went over some distributive properties and some really hard questions that are gonna be on the quiz. Though he didn't give us the answers, it's pretty easy but you really shouldn't mess up like writing down the wrong problem in your equation. So tonight I'm gonna study really really really really hard. Wish me luck!