Graph x = y^2-6y+11
a. In function mode using two functions
b. In parametric mode
I already figured out how to graph it in functions, which is +/-sqrt(x-2)+3. The graph is a sideways parabola in the first quadrant opening to the right.
I have no idea on how to come up with parametric equations for this. Help?
Thanks!
To graph the equation x = y^2 - 6y + 11 in parametric mode, we need to express both x and y in terms of a parameter (often denoted as t). Here's how we can do it:
Let's begin by completing the square for the quadratic equation y^2 - 6y + 11 = 0. We can rewrite it as (y - 3)^2 - 9 + 11 = (y - 3)^2 + 2 = 0.
Now, we can rearrange the equation as follows:
(y - 3)^2 = -2
y - 3 = ±√(-2)
y = 3 ± i√2
Since we can't graph complex numbers on the Cartesian plane, we'll only consider the real part of the equation:
y = 3
Now, let's choose a parameter, t, and express x and y in terms of t:
x = t
y = 3
We have our parametric equations:
x = t
y = 3
To graph these parametric equations, you can assign different values to t and evaluate x and y accordingly. Plot the points (x, y) on the Cartesian plane, and you'll see a line parallel to the x-axis at y = 3.
In summary, the graph of x = y^2 - 6y + 11 in parametric mode consists of a horizontal line at y = 3.