Rewrite the equation in standard form, then find the vertex of the graph of $$ f(x) $$. $$ f(x)=x^{2}-10 x+23 $$. In standard form, $$ f(x)=\square $$ The vertex is at

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1. Given the function  
   $$
   f(x) = x^2 - 10x + 23,
   $$  
   we complete the square.

2. Group the quadratic and linear terms:  
   $$
   x^2 - 10x.
   $$

3. To complete the square, take half of the coefficient of $$x$$ (which is $$-10$$) and square it:  
   $$
   \left(\frac{-10}{2}\right)^2 = 25.
   $$

4. Add and subtract $$25$$ inside the function:  
   $$
   f(x) = \left( x^2 - 10x + 25 \right) + 23 - 25.
   $$

5. Rewrite the perfect square trinomial and simplify the constant terms:  
   $$
   f(x) = (x-5)^2 - 2.
   $$

6. Therefore, the standard form of the function is  
   $$
   f(x) = (x-5)^2 - 2.
   $$

7. The vertex of the parabola given by a function in the form $$ f(x) = (x-h)^2 + k $$ is  
   $$
   (h, k).
   $$  
   Comparing, we have $$ h = 5 $$ and $$ k = -2 $$.

8. Thus, the vertex is at  
   $$
   (5, -2).
   $$
In standard form, $$ f(x) = (x - 5)^2 - 2 $$. The vertex is at (5, -2).

Rewrite the equation in standard form, then find the vertex of the graph of \( f(x) \). \( f(x)=x^{2}-10 x+23 \). In standard form, \( f(x)=\square \) The vertex is at

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