PLS HELP WILL GET MARKED AS BRAINLIST Part A: Explain why the x-coordinates of the points where the graphs of the equations y = - 4x and y = 2(- x + 1) Intersect are the solutions of the equation - 4x = 2(- x + 1) (4 points) Part B: Make tables to find the solution to - 4x = 2(- x + 1) . Take the integer values of between - 2 and 2. (4 points) Part How can you solve the equation - 4x = 2(- x + 1) ) graphically ? (2 points)

Ok, this is quite a bit more work than I am used to but I will try!

Step-by-step explanation:

Ok, seeing we have no hidden value for x  I presume that we are using it as a coordinate plane listing location. Now that we have that settled, that would mean for the first bit on "A" is y= -4x is starting at x=0 on the number line and we are moving it back 4 spaces. Next bit for "A" is y=2(-x+1) seeing that it is right beside it means it is multiplication and we will be using PEMDAS for this one so we can get the best results. We will start with taking -x which automatically makes it -1x because there is no value infront of it and adding 1 because of +1, you can't have -0 you know! Meaning this value is y=0 because anything times zero is, say it with me! ZEEERR! Now for part B, we moved back to "x" but for both parts... hmm well, let's start! -4x=2(-x+1), practically our last question but with multiple "x's". Using same logic, answer is -4x=0, meaning the value of "x" is 0. Turns out last questions "y" variable was an actual variable! Not a piece on the coordinate plane, silly me! Next answer is 0 and after that -4x=2(-x+1) we got the same logic as our last question, almost exact actually! There fore -4x=0! Hope this helped and have a good day! :D

a function could be identified in 3 ways either on a chart,graph, or ordered pairs. first a function is that each point has its own. the rule is that there can be many x's to one y but there can not be 2 x's at different y's. so on a graph when you put a straight line through the y's there should be one dot only to be a function.

step-by-step explanation: