# Ordered Pairs (x,y)

• Jun 11th 2007, 12:56 PM
blueridge
Ordered Pairs (x,y)
For each question below, tell whether the set of ordered pairs (x,y) defined by each equation is a function.

(a) x + 2y^2 = 1

(b) y = (2/x)
• Jun 11th 2007, 01:11 PM
Jhevon
Quote:

Originally Posted by blueridge
For each question below, tell whether the set of ordered pairs (x,y) defined by each equation is a function.

(a) x + 2y^2 = 1

(b) y = (2/x)

hint: we have a function if each x value maps to one and only 1 y value. is this true for both these formulas?
• Jun 11th 2007, 02:25 PM
ThePerfectHacker
Quote:

(a) x + 2y^2 = 1
No. Because $(x,y) = \left( 0, \frac{1}{\sqrt{2}}\right) = \left(0, -\frac{1}{\sqrt{2}}\right)$
Both work. For the same x we get more than one y.
• Jun 11th 2007, 04:00 PM
blueridge
tell me...
What do you mean that both work?

Are these one-to-one functions?

Where did you get those values in terms of (x,y)?
• Jun 11th 2007, 04:04 PM
Jhevon
Quote:

Originally Posted by blueridge
What do you mean that both work?

he found a single value for x that maps to two values of y, so it is not a function, since both y's work for 1 x

Quote:

Are these one-to-one functions?

the first one is not a function at all, the second is a 1-1 function

Quote:

Where did you get those values in terms of (x,y)?
through inspection. he realized that the y is squared, so for any positive y-value we plug in, we will get the same answer as when we plug in the negative of that value