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)
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
+36916 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
