This time we went over perpendicular lines and parallel lines and how they work.

Perpendicular lines can be represented by a 90* angle. It’s also when one fraction is a reciprocal of the other. like 3/4 becomes -4/3

Parallel lines can be described by two equal slopes like. 3/4 and 6/8

this time is how we calculated the Distance over something. If we want the slope of A and B

we take the coordinates (0,3) (4,6)  and we use the equation m=y2-y1/x2-x1 or m=y1-y2/x1-x2 so get the slope=3/4

Then take the equation y=mx+b and put the slope in the equation: y=3/4x+b.

After successfully completing these steps you choose one of the two coordinates and replace the y and x with them.

6=3/4×4+b

which gives us the y-intercept/b

3=b

and there we have everything.

This week we went through Slope-Y intercept form, general form, Point slope form. In the picture we began with Point slope then continued with Slope-Y intercept form then finished with General form.

Systems of Linear Equations: Naming on Desmos

Yesterday we made names on desmos with points to create names. In the image above you can see my initials DY starting from x=0 and y=0

Now we magically used General form, Slope-y intercept form, and point slope form to re create the initials of our name. The top 4 is the D and the bottom 3 is the Y

Function Notation 

So in Function notation we use the input to find the output by using this equation. There are multiple ways to represent this including mapping notation.

So for say if the Input was 1 we would replace the x with 1. so 3*1=3

3+1=4 so when x=1 y=4

So every time X climbs, Y can climb with it. Or decrease depending on the question.

Functions and Relations

Today we were introduced into the universe of linear equations.

We learn that input is x and output is y

and several different possible use of an equation

f : x–> (0)x-1

f(x)=(0)x-1

As replacing the x in front of the f with a number then substituting the x in the equation gives us a problem to solve.

Solving this problem gives us Y

Polynomials: Factoring

Today we learned that as we can develop the question we must be able to do it backwards.

In this case we factored the question back to a simpler state.  Solving this question is focusing on the 2x and the -5

I factor starting with 2 multiplying 5 making 10, then from 10 we determine how to get 9; which in this case is only 1 and 10 as its factors

so we put down (2x-1)(2x+10)

then simplying the 2 and the 10

=(2x-1)(x+5)