Question 825613
A standard quadratic equation like y = x^2 is a U-shaped curve that is centred on the origin, has its axis of symmetry as the line x = 0, and has its turning point at the origin also, i.e. where y = 0.
If we translate this graph about the x-y plane such that its origin is now at the coords (h,k), then the new equation of the curve is,
(y-k) = (x-h)^2
And here the axis of symmetry is the line x = h and the bottom of the curve (its turning point) is on the line y =  k.
Our graph is y = x^2 - 6x + 10
completing the square on the rhs gives us,
y = x^2 - 6x + 9 + 1
y = (x - 3)^2 + 1
(y - 1) = (x - 3)^2
Here (h,k) = (3,1).
So the origin of this curve is the point (x,y) = (3,1).
Thus the y-coordinate of its turning point is y = 1.