Let f(x)=x^2+bx+27. Find the value(s) of b so that f(x) has minimum value of 2.

Answer 1

any parabola y=ax^2 + bx + c has a minumum when x = -b/2a

so, f(x) has a minimum when x = -b/2

f(-b/2) = (b^2/4) + b(-b/2) + 27

so, if f(x) = 2 then

b^2/4 + -b^2/2 + 27 = 2
-b^2/4 + 25 = 0
b^2 = 100
b = +/- 10

That translates into
(x+5)^2 + 2
or
(x-5)^2 + 2

Both have a min at y=2.