This week we looked at liner graphs relations, and smothign that I have taken away fomr the lesson is, point slope form. I find this is one of the easiest forms to find y intercepts, and plotting points. Let me show you.

Say we had a been given a random set of points and a slope, like; m=4 (2,3) **m=slope**

Now the formula for point slope form is, m(x(-+)x1)=y(-+)y1; so if we substitute in our values, making sure that we switch the integer value of the plotted points (because we are trying to get to the next point/y intercept):

4(x-2)=y-3 –> this is point slope form, Now if we were to find the y intercept then we would need to do some algebra; and or first step would be to get y alone, by moving the 3 over the the rest of the equation by reciprocating its integer value:

4(x-2)+3=y

our next step would be to get rid of the brackets by using distributive property:

4(x) 4(-2) +3 = y

now our equation looks like this:

4x-8+3=y

All we have left to do is simplify:

4x-5=y

You probably have noticed, wait a minute….. now our equation is in intercept slope form! Well you are right. This is why starting with point slope form is easier to fill in the blanks, although if it is easier for you, then do algebra by substituting in the values in intercept slope form to find the missing values. you can also turn the point slope form/intercept slope form in to general form, but that’s for next week.