Question 1120176
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You might want to look at the lesson tutor @ikleyn suggested in her response.  Completing the square is a skill you might need in a lot of different types of math problems.<br>
In case they might help your understanding of the process, let me add a few notes of explanation to the solution as she showed it....<br>
{{{2x^2-8x+11}}}<br>
Ignore the constant term for the moment; you need to complete the square in the variable x.
{{{(2x^2-8x)+11}}}<br>
To complete the square, you need to factor out the leading coefficient:
{{{2(x^2-4x)+11}}}<br>
Think of the pattern for squaring a binomial: (x+a)^2 = x^2+2ax+a^2.  The coefficient of the middle term in the product is twice the constant term in the binomial.  So to complete a square you need to take half the coefficient of the linear term and square it.<br>
In this example, the coefficient of the linear term is -4; half of that squared is (-2)^2 = 4.  You need to complete the square in the parentheses by adding 4.<br>
Note that you have added 4 inside the parentheses, so you have added 2*4=8 to the expression as a whole.  To balance that, you need to subtract 8 from the expression (outside of the parentheses).
{{{2(x^2-4x+4)+11-8}}}<br>
Now write the trinomial as a binomial squared, and simplify the rest of the expression.
{{{2(x-2)^2+3}}}<br>
The square of a number is always 0 or positive, so the minimum value of the expression will be when (x-2) is zero -- i.e., when x=2.  And the value of the expression when x=2 is 2(0^2)+3 = 3.  So the vertex of the graph is (2,3).