c'mon, this is just Algebra I. The vertex is at x = -b/2a = 1
So the minimum is at (1,f(1)) - (1,-6+k)
this is easy to see, since
f(x) = 6(x-1)^2 + k-6
What is the minimum value of the given function? 𝑓(𝑥)=6𝑥^2 −12𝑥+𝑘, 0≤𝑥≤10, 𝑘 is a constant.
(A)𝑓(0) (B) 𝑓(10) (C) 𝑓(0) + 𝑘 (D) 𝑓(1) + 𝑘 (E) none of the above.
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