Algebra ( Period 5)

Chapter 3-6 Proportional and Nonproportional Relationships

This is just real world review of concepts we’ve already covered… comparing and contrasting the two types of linear relationships

SAME: both are linear—meaning they graph as lines
both are diagonal
both have a constant rate of change or slope that can be found by finding the rise/run or the difference of the y’s over the difference of the x’s

DIFFERENT:
proportional relationships go through the origin (0,0) and nonproportional do NOT

Nonproportional have a y-intercept other than 0

Proportional relationships: YOU can take any point and divide the y/x and it will equal the same value as diving any other y/x. This value  is the slope—which is the constant rate of change  VS Nonproportional relationships when you divide the y/x of a point it will NOT equal the y/x of another point. This value is NOT the slope and is NOT the CONSTANT RATE OF CHANGE

TO find the equation for anonproportional relationship
This isn’t as easy as f(x) = kx because it does not go through (0,0)

You will need to find the y intercept (0,y)

Say you find the rate of change or slope is 3 for the following 2 points

(2, 12) and (4, 18)

You can graph these two points and count the slope down to the y intercept

You can find a missing number that will make the equation work

y = 3x + ?  will make ( 2, 12) work in the equation

Plug in( 2, 12) to find b

12 = 3(2) + b

12 = 6 +b

 b = 6

You can try it with other points as well…

 It still works b = 6

So the non-proportional equation is y = 3x + 6

6 is the y intercept on the graph