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.
I have no idea on how to come up with parametric equations for this. Help?
Thanks!
Certainly! To express the equation x = y^2 - 6y + 11 in parametric mode, we can introduce a new variable, say t, and rewrite the equation as a system of parametric equations. Here's how you can do it:
Step 1: Solve the equation for y in terms of x:
x = y^2 - 6y + 11
y^2 - 6y = x - 11
Completing the square, we have:
(y - 3)^2 = x - 11 + 9
(y - 3)^2 = x - 2
y - 3 = ±√(x - 2)
y = ±√(x - 2) + 3
Step 2: Express x and y in terms of the parameter t:
To do this, we let x = t and substitute it into the equation for y:
y = ±√(t - 2) + 3
So the parametric equations for the graph x = y^2 - 6y + 11 are:
x = t
y = ±√(t - 2) + 3
These parametric equations represent the same graph as the function form, but they provide a different way of representing the relationship between x and y.