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

what the question so we can you out

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: