Identify the domain and range, then state whether the relation is a function:

Unlike the problem we looked at before, we have individual points here. We can't assume that the stuff in between is also included in this relationship. The domain is x = -3, 1, and 3. The range is y = -5, 2, 7, and 9.

Is this a function? No way José is this a function. We have x = 3 doing double duty with y = 7 and 9. That's no good. If someone asked us to find f(3), we wouldn't know which y to give them.

Identify the domain and range, then state whether the relation is a function:

The domain is x = -3, 1, 3, and 5. The range is y = -5, 2, and 7. The relation is a function, because each input has exactly one output. We never said that each output could happen only once. The only restriction is one x enters the equation and then one y leaves.

Graph f(x) = 2x + 1 by plugging in values for x.

Don't let that f(x) fool you. It's just y going by an assumed name. Apparently Seymour Butz was already taken.
Let's make a table for our plugged-in values. Just a few will do.



Graph the points and connect the dots.



Our domain isn't restricted, so the function keeps going and going and going.