Ordered Pairs (x, y)

• Jan 18th 2007, 01:09 PM
symmetry
Ordered Pairs (x, y)
For questions a - b, tell whether the set of ordered pairs (x, y) defined by each equation is a function.

(a) y = x^3 - 3x

(b) y = + - sqrt{1 - 2x}

(c) x + 2y^2 = 1

How can I tell simply by looking at each equation?
• Jan 18th 2007, 02:22 PM
topsquark
Quote:

Originally Posted by symmetry
For questions a - b, tell whether the set of ordered pairs (x, y) defined by each equation is a function.

(a) y = x^3 - 3x

(b) y = + - sqrt{1 - 2x}

(c) x + 2y^2 = 1

How can I tell simply by looking at each equation?

I wouldn't look at the equation, I'd look at the graph. Do you know what the "vertical line test" is?

Graph each function. If, when you pass a vertical line through any portion of the graph, the line intersects your function in at most one point, the graph defines a function.

We can easily see the b and c are not functions, whereas a is.

-Dan
• Jan 18th 2007, 05:35 PM
symmetry
ok
Yes, I know what the vertical and horizontal line tests are all about.

Okay...so it is better to graph each function and apply the vertical line test to determine if it is a function.

Thanks!
• Jan 18th 2007, 08:22 PM
CaptainBlack
Quote:

Originally Posted by symmetry
For questions a - b, tell whether the set of ordered pairs (x, y) defined by each equation is a function.

(a) y = x^3 - 3x

(b) y = + - sqrt{1 - 2x}

(c) x + 2y^2 = 1

How can I tell simply by looking at each equation?

Is there a single y corresponding to each x where the relation is defined?
If the answer is no its not a function.

Thus (b) and (c) fail to be functions, while (a) does not fail this for any
real x, so it defines a function on R to itself.

RonL