# Linear functions

Printable View

• Aug 29th 2007, 04:03 PM
Ash
Linear functions
What are the steps to solving this problem???

- Which could be linear functions? Explain.-

Relation 1:
x = 1 2 3 3 4
y= 3 4 5 6 7

Relation 2:
x= 5 5 5 5 5
y= 27 24 21 18 15

Relation 3:
x= 0 1 2 3 4
y= 50 45 40 35 30

Relation 4:
x= 3 4 5 6 7
y= 11 13 17 25 40
• Aug 29th 2007, 04:18 PM
ThePerfectHacker
Quote:

Originally Posted by Ash
What are the steps to solving this problem???

- Which could be linear functions? Explain.-

Relation 1:
x = 1 2 3 3 4
y= 3 4 5 6 7

You need to increase (or decrease) by the same amount as you increase x by 1 unit. Do you?
Quote:

Relation 2:
x= 5 5 5 5 5
y= 27 24 21 18 15
This is not even a function, why?

Quote:

Relation 3:
x= 0 1 2 3 4
y= 50 45 40 35 30
You need to increase (or decrease) by the same amount as you keep increase x by 1 unit. Do you?

Quote:

Relation 4:
x= 3 4 5 6 7
y= 11 13 17 25 40
Same idea.
• Aug 29th 2007, 04:18 PM
Plato
In any function no two ordered pairs have the same second term.
Why does that rule out the first two?

In order to be linear the collection of pairs must have a constant relation.
If $\displaystyle \left( {x_1 ,y_1 } \right),\;\left( {x_2 ,y_2 } \right),\;\left( {x_3 ,y_3 } \right)$ are all pairs in the function then it must be true that
$\displaystyle \frac{{y_1 - y_2 }}{{x_1 - x_2 }} = \frac{{y_3 - y_2 }}{{x_3 - x_2 }} = \frac{{y_1 - y_3 }}{{x_1 - x_3 }}$.
• Sep 3rd 2007, 06:06 PM
Ash
Thanks for the help. Can you check to see if my conclusions are correct.

I know relation 1 and 2 are not functions because in both cases the input leads to two or more outputs.

Relation 3 is a (linear) function because it increases (or decreases) by the same amount- plus, the slope and y-intercept are constant.

I'm guessing relation 4 is a function but, I gather it's not a (linear) function because, when I calculated the points, the slopes came out to be different.
• Sep 3rd 2007, 06:35 PM
ThePerfectHacker
It seems correct.