Methods for finding the minimum coordinate of an equation (any shortcuts?) I want to know if there's a quicker method to find the minimum coordinates of a quadratic equation other than the one described below.
Some background: 
I was reviewing "Completing Squares and Inequalities" when I came to this inequality: $ y \leq x^2 - 2x +2$. This ends up being $(x-1)^2 + 1$ 
I now want to find the minimum coordinates of this expression. The vertex can be found with $\frac{-b}{2a}$ and my current rationale is the following:
Premise I: $a > 0$, then this equation has a minimum
Premise II: Vertex can be found with $[\frac{-b}{2a}, f(\frac{-b}{2a})]$


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*Find the vertex: $\frac{-b}{2a} = \frac{-(-2)}{2(1)} = \frac{2}{2} = 1$

*Then replace that in the original equation:
$(1)^2 - 2(1) +2 = 1 - 2 + 2 = 1$
Alternatively I can make use of the "Vertex form" $a(x - h)^2 + k$ and get that from the "completed square" $(x - 1)^2 + 1$ so we know that the Vertex is 1 and repeat the two steps above.
Other than this method, do we have a shortcut for finding the minimum coordinates? The book tells it is $(1,1)$, but doesn't show the method.

You can "see" the minimum from an expression like: $$(x-1)^2+1$$
Since $(x-1)^2$ is a square which is always $\ge 0$, the minimum is $1$.

In general for an expression like
$$a(x\pm h)^2 + k$$
the minimum is $k$ when $a\gt 0$
