1. ## 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)

2. 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?

3. (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.

4. ## 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)?

5. 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

Are these one-to-one functions?
the first one is not a function at all, the second is a 1-1 function

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