Quadratics?

Question

3y^2 - x - 6y + 5 = 0

Can anyone tell me what the vertex is please?

Answer 1

First you move some variables around to get:
3y^2 - 6y +5 = x
This is a parabola that opens to the "right", since the quadratic is in y, and the quadratic term is possible. The vertex is where the minimum value of the parabola is: If you have a ay^2 + by + c = x
The vertex value of the y-coordinate is given by
2ay + b =0 or y= -b/(2a). In this case y=1 and x=2 so the coordinates are (2,1).

Answer 2

Hi,

The vertex is at(2,1).

If you complete the square,

x = 3y^2 - 6y + 5

x = 3(y^2 - 2y ) + 5

x = 3(y^2 - 2y +1 ) + 5 - 3

x = 3(y - 1)^2 + 2

This is in vertex form, x = a(y - k)^2 + h

Since the vertex is at (h,k), it's at (2,1).


I hope that helps!! :-)