Question 962422
f(x)=(x-2)(x-3)=x^2-5x+6

To find the x coordinate of the vertex, there is a formula: (-b/2a). Since I multiplied the x-2 and x-3 to get the quadratic equation into ax^2+bx+c form (standard form). In this case, a=1 and b=-5.
Plugging these values into the formula (-b/2a), we get 5/2, or 2.5. We can substitute 2.5 back into your equation to find what y is.
(2.5-2)(2.5-3)= 0.5(-0.5)=-0.25
Vertex: (2.5, -0.25)

Since the axis of symmetry is a line going through the vertex vertically, it has to pass through the x axis and therefore is the x coordinate of your vertex. x=2.5

The y-intercept is when the x coordinate equals 0, so we can substitute 0 for x in your equation. 
(0-2)(0-3)=-2*-3=6
y-intercept= 6

This parabola has a minimum value. It starts at the vertex (2.5,-0.25), and opens upward. If a>0, it has a minimum. If a<0, it has a maximum. In your function, a=1, so it opens upward and the vertex is the minimum point.