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For what constant k is 1 the minimum value of the quadratic 3x^2 - 12x + k over all real values of x? (x cannot be nonreal)

 Oct 24, 2020
 #1
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we have 3x^2-12x+k and one of the roots has to be (x-1) or (3x-1). (x-1) is more convinent so we'll use that. With this, we can add 3 to -12 because -1 will be multiplied by 3, and then we get -9x remaining. So we get the equation is (x-1)(3x-9) and k is 9

 Oct 24, 2020
 #3
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Sorry instead of x-1 being more convinient, I meant that it had to be x-1 because if it was 3x-1 it would be 1/3. So it would be (3x-9)(x-1)

However, using 3x-3 we would get (3x-3)(x-3)

Guest Oct 24, 2020
 #2
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This is a bowl shaped parabola

minimum occurs at the vertex

'x' at the vertex can be found by   - b /2a =  - -12/6 = 2

 

sub in 2 for x

3(2^2) -12(2) + k = 1              1 is the minimum value requested in the Q

12   - 24   + k = 1

k = 13

 

Here is a graph:   (you can see the minumum value is 1 when k = 13)

https://www.desmos.com/calculator/ar7evacrbn

 Oct 24, 2020
edited by ElectricPavlov  Oct 24, 2020

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