Express the function f(x) =3x^2-6x-18 in the form a(x + h)^2=k, where a, h and k are constants. Hence state the : (i) the minimum value of f(x).
(ii) the value of x for which f (x) is a minimum

The minimum value of f(x) is -21 and it occurs at x = 1

Step-by-step explanation:

f(x) =3x^2-6x-18

Factor out the greatest common factor out of the first two terms

f(x) =3(x^2-2x)-18

Complete the square

-2x/2 =-1  (-1)^2 = 1

Add 1  (But remember the 3 out front so we are really adding 3  so we need to subtract 3 to remain balanced)

f(x) = 3(x^2 -2x+1) -3 -18

f(x) = 3(x-1)^2 -21

This is vertex form

f(x) = a(x-h)^2 +k  where (h,k) is the vertex and a is a constant

The vertex is (1,-21)  

Since a > 0 this opens upward and the vertex is a minimum

The minimum value of f(x) is -21 and it occurs at x = 1

answer: 12

step-by-step explanation:

because yes