Question 1026782
{{{2s^2-8s+19}}} is a quadratic polynomial.
Quadratic functions such as {{{f(x)=ax^2+bx+c}}} with {{a<>0}}} ,
have a maximum (if {{{a<0}}} ),
or a minimum (if {{{a>0}}} ).
So, since the coefficient of {{{s^2}}} is {{{2>0}}} ,
{{{2s^2-8s+19}}} has a minimum value}}} .
 
How could you find that minimum value?
You could remember and apply a recipe handed down by your algebra 2 teacher.
Alternatively, you could use your knowledge of algebra, and your ability to think.
 
APPLYING MEMORIZED FORMULAS/RECIPES:
A quadratic function {{f(x)=ax^2+bx+c}}} with {{{a<0}}} has a minimum for {{{X=(-b)/"2 a"}}} .
In the case of {{{2s^2-8s+19}}} ,
the variable is {{{s}}} rather than {{{x}}} , {{{a=2}}} and {{{b=-8}}} ,
so the minimum happens for {{{s=(-(-8))/(2*2)=8/4=2}}} ,
and for {{{s=2}}} , {{{2s^2-8s+19=2*2^2-8*2+19=2*4-16+19=8-16+19=highlight(11)}}} .
 
IF YOU DID NOT MEMORIZE THAT FORMULA,
you have to reason your way to the answer.
{{{2s^2-8s=2(s^2-4s)}}} may remind you of {{{s^2-4s+4=(s-2)^2}}} ,
and {{{2(s-2)^2=2(s^2-4s+4)=2s^2-8s+8}}} .
So, you can complete the square:
{{{2s^2-8s+19=2s^2-8s+8+19-8=2(s-2)^2+19-8=2(s-2)^2+11}}} .
You know that the square of any real number is non-negative,
so {{{(s-2)^2>=0}}} , and {{{2s^2-8s+19=2(s-2)^2+11>=0+11=11}}} ,
which means the the smallest possible value of {{{2s^2-8s+19}}} is {{{highlight(11)}}} .