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)

Guest Oct 24, 2020

#1**0 **

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

Guest Oct 24, 2020

#2**+1 **

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

ElectricPavlov Oct 24, 2020