R rachael724 Junior Member Joined Sep 14, 2005 Messages 95 Oct 5, 2005 #1 What is the maximum value for f(x) = -3x2 + 12x - 1 a. 11 b. 12 c. 13 d. 14 e. 15 f. 16 g. 17 h. 18 i. none of these I am not too well educated on maxium value.

What is the maximum value for f(x) = -3x2 + 12x - 1 a. 11 b. 12 c. 13 d. 14 e. 15 f. 16 g. 17 h. 18 i. none of these I am not too well educated on maxium value.

T TchrWill Full Member Joined Jul 7, 2005 Messages 856 Oct 5, 2005 #2 f(x) = -3x^2 + 12x - 1 Taking the first derivitive and setting equal to 0, yields df(x)/dx = -6x + 12 = 0 making x = -12/-6 = 2. Therefore, the maximum value of f(x) is -3(2^2) + 12(2) - 1 = 11

f(x) = -3x^2 + 12x - 1 Taking the first derivitive and setting equal to 0, yields df(x)/dx = -6x + 12 = 0 making x = -12/-6 = 2. Therefore, the maximum value of f(x) is -3(2^2) + 12(2) - 1 = 11

T ting Junior Member Joined Sep 18, 2005 Messages 87 Oct 5, 2005 #3 Find the vertex. Could complete the square. -3x^2 + 12x -1 Divide through by -3 =-3[(x^2 - 4x) + (1/3)] =-3[(x - 2)^2 - (-2)^2 +(1/3)] =-3[(x - 2)^2 - 4 +(1/3)] =-3[(x - 2)^2 - (12/3) +(1/3)] =-3[(x - 2)^2 - (11/3)] =-3(x - 2)^2 + 11 What is the y coordinate of the vertex?

Find the vertex. Could complete the square. -3x^2 + 12x -1 Divide through by -3 =-3[(x^2 - 4x) + (1/3)] =-3[(x - 2)^2 - (-2)^2 +(1/3)] =-3[(x - 2)^2 - 4 +(1/3)] =-3[(x - 2)^2 - (12/3) +(1/3)] =-3[(x - 2)^2 - (11/3)] =-3(x - 2)^2 + 11 What is the y coordinate of the vertex?