Question 1040891
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If s is a real number, then what is the smallest possible value of 2s^2 - 8s + 19?
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<pre>
1.  Firmly memorize this:

    A quadratic function  y = {{{ax^2 + bx + c}}}  has a minimum/maximum at  x = {{{-b/2a}}}.     (1)

    To find this maximum/minimum value, substitute  x = {{{-b/2a}}}  into the quadratic function. 
    Then min/max, after calculations, is  {{{-(b^2-4ac)/4a}}}.                              (2)

    Next,  if  a > 0  then the parabola is open up and the min/max is the minimum.
           If  a < 0  then the parabola is open down and the min/max is the maximum.


2.  In your case,  the minimum is at s = {{{-(-8)/(2*2)}}} = 2.
    To calculate {{{y[min]}}}, substitute s = 2 into the quadratic function.  It will produce the same value as (2), but the calculations are easier.

    {{{y[min]}}} = {{{2*2^2 -8*2 +19}}} = 8 - 16 + 19 = 11.
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See the lesson <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Who-is-who-in-quadratic-equations.lesson>Who is who in quadratic equations</A> in this site.


<pre>
Next, memorize this:


     It doesn't matter which letter of the English alphabet is used in the parabola equation as a variable. 
     To find a min/max and a vertex coordinates, always use the formula (1). It works ALWAYS.

     After finding x, calculate the value of a min/max by substituting the found value  of "x" into the parabola equation.
</pre>
And the last notice.


There is no need to use so many symbols "$" as you do.
We understand clearly your formulas without it.