Relations and Functions

Relations and Functions

In N Out Burger

If you and your friend walk into In-N-Out Burger and you both order a Double-Double, will you both pay the same price?

There's only one price for each item, right?

• If two items had the same price, that would be OK, right? Like, on a Dollar Menu?

• But what if teh same item had two different prices?

Open Ended

My football people: if I say Green Bay, you say:

Open Ended

If I say New England, you say:

Open Ended

If I say Chicago, you say:

Open Ended

If I say New York, you say:

That last question, that's a problem.

In algebra, anyway. I only want to have one "output" for any one "input".

That's how we define a "function".

Any set of ordered pairs is a "relation". But in a "function" there is only one output value for any given input value.

Multiple Choice

Is the relation

 $\left\{\left(0,\ 1\right),\ \left(3,\ 4\right),\ \left(5,\ 6\right),\ \left(7,\ 8\right)\right\}$  a function?

• Yes

• No

Multiple Choice

How about

 $\left\{\left(0\ ,\ 1\right),\ \left(3,\ 5\right),\ \left(0,\ 7\right),\ \left(8,\ 9\right)\right\}$  ?

• Yes

• No

Open Ended

Explain your answer to the previous question.

Multiple Choice

Is the relation

 $\left\{\left(-2,\ 4\right),\ \left(-1,\ 1\right),\ \left(0\ ,\ 0\right),\ \left(1,\ 1\right),\ \left(2,\ 4\right)\right\}$  a function?

• Yes

• No

It's fine for more than one x value to go to the same y.

You just can't have the same x go with more than one y.

Multiple Choice

What value could I enter for x to make this relation not a function?

 $\left\{\left(1,\ 4\right),\ \left(-2,\ 3\right),\ \left(x,\ 5\right),\ \left(7,\ -2\right)\right\}$  

• -2

• 5

• 3

• 0

Open Ended

Explain your answer to the previous question.