SOLUTION: 2. Determine whether the relation {(–2, 3), (–5, 6), (3, 0), (1, 1)} is a function 3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a fu

Algebra ->  Functions -> SOLUTION: 2. Determine whether the relation {(–2, 3), (–5, 6), (3, 0), (1, 1)} is a function 3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a fu      Log On


   



Question 495115: 2. Determine whether the relation {(–2, 3), (–5, 6), (3, 0), (1, 1)} is a function
3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function.
Find ƒ(–5) for each function.
5. ƒ(x) = 5x + 35
6. ƒ(x) = x² – x


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
2. Determine whether the relation {(–2, 3), (–5, 6), (3, 0), (1, 1)} is a function
We look at the first coordinates -2, -5, 3, and 1. None of those are
the same, so it is function.

3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function.
We look at the first coordinates -4, 1, 0, and -4. Two of those are
the same -4, so it is NOT a function.  To make it into a function we can
delete either (-4, 2) or (-4, 6).  You can put either of these as the
solution:

{(1, 6), (0, 0), (–4, 6)} or {(–4, 2), (1, 6), (0, 0)}




Find ƒ(–5) for each function.
5. ƒ(x) = 5x + 35
Everywhere you see an x, replace it by (-5)

     ƒ(-5) = 5(-5) + 35

Then simplify the right side:

     ƒ(-5) = -25 + 35

     ƒ(-5) = 10


6. ƒ(x) = x² – x
Everywhere you see an x, replace it by (-5)

     ƒ(-5) = (-5)² - (-5)

Then simplify the right side:

     ƒ(-5) = 25 + 5

     ƒ(-5) = 30
 
Edwin