Identify the function type:
Quadratic: (y=ax^2+bx+c)
Circle: ((x-h)^2+(y-k)^2=r^2)
Parabola given in vertex form: (y=a(x-h)^2+k)
For a quadratic (y=ax^2+bx+c):
Vertex (x)-coordinate: (x_v=-frac{b}{2a})
Vertex (y)-coordinate: (y_v=f(x_v)=aleft(-frac{b}{2a}right)^2+bleft(-frac{b}{2a}right)+c)
Vertex point: ((x_v, y_v))
Using the completing-the-square method for (y=ax^2+bx+c):
Rewrite (y=aleft(x+frac{b}{2a}right)^2+left(c-frac{b^2}{4a}right))
Vertex: (left(-frac{b}{2a},,c-frac{b^2}{4a}right))
If the quadratic is already in vertex form (y=a(x-h)^2+k):
Vertex: ((h,k))
For a circle ((x-h)^2+(y-k)^2=r^2):
Vertex (center): ((h,k))
For a parabola with factored form (y=a(x-r_1)(x-r_2)):
Axis of symmetry (vertex (x)-coordinate): (x_v=frac{r_1+r_2}{2})
Vertex (y)-coordinate: (y_v=f(x_v))
Vertex point: ((x_v, y_v))
If given two points on a quadratic and you need the vertex:
Find the quadratic equation first (e.g., solve for (a,b,c))
Then use (x_v=-frac{b}{2a}) and (y_v=f(x_v))