**Parametric Equations** are pairs of continuous functions
that define the x and y coordinates of a point in the plane bases
on a third parameter.

In other words, there is one equation to find the x-value and a different equation to determine the y-value. This two values are then a point (x, y) that can be plotted in the coordinate plane.

**The general form is: **

x(t) = an expression that gives the x-value and

y(t) = an expression that gives the y-value.

where the value t to be substituted into the two functions is called the parameter.

**For example: **

x(t) = t+ 1 and y(t) = t -3. (these are parametric equations)

Now, we can find points (x, y) based on different values of t. This will then give some kind of graph.

So, let's look at the values of t from -3 to 3.

tx(t) = t + 1y(t) = t - 3(x, y)-3 -3 +1 = -2 -3 - 3 =- 6 (-2, -6) -2 -2 +1 = -1 -2 - 3 =- 5 (-1, -5) -1 -1 +1 = 0 -1 - 3 =- 4 (0, -4) 0 0 +1 = 1 0 - 3 =- 3 (1, -3) 1 1 +1 = 2 1 - 3 =- 2 (2, -2) 2 2 +1 = 3 2 - 3 =- 1 (3, -1) 3 3 +1 = 4 3 - 3 =0 (4, 0) We now have seven points to graph and get the following.

Notice anything? Well first of all the points are linear. Is there a reason why? The answer to that is yes. I stems from the parametiric equations. Both of them are the same degree, so the graph will be linear. When one has degree one and the other has degree greater than 1, the graph of the parametric equations takes the shape of the individual parametric equation with the highest degree. When they both have degee greater than 1, many different scenarios can occur.

10 Point Bonus:What happens when both have degree greater than 1 and are odd. Make sure to have graphs to illustrate your theory.

Based on this, what would one expect the shape of the graph of parametric equations where one has degree one and the the other has degree two? Let's investigate and see what happens.

**For Example:**

x(t) = t + 1 and y(t) = t^2. Find the values for the range of t from -2 to 2.

xx(t) = t + 1y(t) = t^2(x, y)-2 -2 + 1 = -1 (-2)^2 = 4 (-1, 4) -1 -1 + 1 = 0 (-1)^2 = 1 (0, 1) 0 0 + 1 = 1 (0)^2 = 0 (1, 0) 1 1 + 1 = 2 1^2 = 1 (1, 2) 2 2 + 1 = 3 2^2 = 4 (3, 4) We now plot the points and get a graph that looks like.

and if we connect the points we get a graph that looks like the following

So, if you guessed that the graph would be a parabola since second degree equations are parabolas. You were correct.

Now, based on what you know, what do you think would happen if both parametric equations are second degree equations? It should be linear since they have the same degree.

Based on this example, that statemet that equal degree parametric equations are linear seems to hold.

What about a first degree and a third degree equation? Based on the earlier statement this should take the shape of a third degree equation which is an S. Look at the example below to see if that holds.

Again the statement appears to hold.

Now, you should be able to solve, graph, and make conjectures about parametric equations.