Methods for finding the minimum coordinate of an equation (any shortcuts?)

Question

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 $[\frac{-b}{2a}, f(\frac{-b}{2a})]$

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

2. 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.

Answer 1

Derivative is also one of tool,

$\frac{dy}{dx}=2x-2 =0$

At $x=1$:

$\frac{d^2y}{dx^2} = 2 >0$

We get min. At $x=1$

Answer 2

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 $1$.

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