# Domain and Range of the parametric equation

• Mar 25th 2009, 03:33 PM
oceanmd
Domain and Range of the parametric equation
Please check whether the following problem s solved correctly. Thank you.
Graph the following parametric equation. Identify the domain and range. Eliminate the parameter.

x=2+q
y=4-2q
Constructed a table:
q: -2, 0, 1, 2
x: 0, 2, 3, 4
y: 8, 4, 2, 0

The raph s a line. Doman: (-infnity, +infnity), Range: (-infinity, + infinity)
After I elminated the parameter, got the equaton: y=-2x+8

Thank you very much for your help.
• Mar 25th 2009, 04:28 PM
Greengoblin
You have a function $f(q)=(x(q),y(q))=(2+q,4-2q)$ whose domain is the real numbers, and whose range is a subset (or equal to) the plane, $\mathbb{R}^2$.

Since x(q)=2+q, then the range of x(q) is the real numbers, $\mathbb{R}$, and since y(q)=4-2q the range of y(q) is also $\mathbb{R}$, therefore, since the function is parametric, the range of the function is the cartesian product of these two ranges, $\mathbb{R}^2$.

so the function is $f:\mathbb{R}\to\mathbb{R}^2$
• Mar 26th 2009, 06:14 AM
stapel
Quote:

Originally Posted by oceanmd
Please check whether the following problem s solved correctly. Thank you.
Graph the following parametric equation. Identify the domain and range. Eliminate the parameter.

x=2+q
y=4-2q

Since x = 2 + q, then q = x - 2. Thus y = 4 - 2(x - 2) = 4 - 2x + 4 = 8 - 2x.

This is a straight line, defined for all x and eventually "covering" all y.

I agree with your solution. :D
• Mar 26th 2009, 09:17 AM
oceanmd
Thank you
stapel,

Thank you.