SOLUTION: If s is a real number, then what is the smallest possible value of 2s^2 - 8s + 19?

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Question 1026782: If s is a real number, then what is the smallest possible value of 2s^2 - 8s + 19?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
2s%5E2-8s%2B19 is a quadratic polynomial.
Quadratic functions such as f%28x%29=ax%5E2%2Bbx%2Bc with {{a<>0}}} ,
have a maximum (if a%3C0 ),
or a minimum (if a%3E0 ).
So, since the coefficient of s%5E2 is 2%3E0 ,
2s%5E2-8s%2B19 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%3C0 has a minimum for X=%28-b%29%2F%222+a%22 .
In the case of 2s%5E2-8s%2B19 ,
the variable is s rather than x , a=2 and b=-8 ,
so the minimum happens for s=%28-%28-8%29%29%2F%282%2A2%29=8%2F4=2 ,
and for s=2 , 2s%5E2-8s%2B19=2%2A2%5E2-8%2A2%2B19=2%2A4-16%2B19=8-16%2B19=highlight%2811%29 .

IF YOU DID NOT MEMORIZE THAT FORMULA,
you have to reason your way to the answer.
2s%5E2-8s=2%28s%5E2-4s%29 may remind you of s%5E2-4s%2B4=%28s-2%29%5E2 ,
and 2%28s-2%29%5E2=2%28s%5E2-4s%2B4%29=2s%5E2-8s%2B8 .
So, you can complete the square:
2s%5E2-8s%2B19=2s%5E2-8s%2B8%2B19-8=2%28s-2%29%5E2%2B19-8=2%28s-2%29%5E2%2B11 .
You know that the square of any real number is non-negative,
so %28s-2%29%5E2%3E=0 , and 2s%5E2-8s%2B19=2%28s-2%29%5E2%2B11%3E=0%2B11=11 ,
which means the the smallest possible value of 2s%5E2-8s%2B19 is highlight%2811%29 .