The function defined by y = 3x^2 - 18x is that of a parabola as shown below:

This is a parabola that opens upwards. As a result it is not possible to determine the maximum value of y.

One may determine the minimum value of y for y = 3x^2 - 18x.

Determine the first derivative y'. This is given by y' = 6x - 18. Solve the equation y' = 0.

6x - 18 = 0

x = 3

At x = 3, y = 3*9 - 18*3 = -27

The minimum value of y for y = 3x^2 - 18x is y = -27. The maximum value of y cannot be determined here.

By using the **maxima and minima procedure** we can find the solution.

as we need to find the maximum or minimun point it is obtained at

`( del y )/(del x) = 0`

ie the partial difference of `y = 0`

Now given

`y = 3x^2 - 18x`

Applying the **partial differentiation** with respect to x

we get

`( del y )/(del x) = 6*x - 18`

As we need to get the max point we need to apply ( del y )/(del x) = 0

so,

=> `0= 6*x - 18`

=> 6*x =18

=> `x= 3`

so at `x=3` y is **maximum or minimum** so the y value when` x= 3` is

`y= 3(3)^2 - 18(3) = - 27`

as the value of y is negative so , it does not have a maximum point but the most minimum value ie `y= -27`

The general form of a quadratic is ax^2+bx+c=0. If "a" is a negative number, the graph has a maximum. In your problem a=3, so it has a minimum, but not a maximum.

The minimum is the vertex. To find the x-value of the vertex we use the formula x=(-b)/2a. For your problem, a=3 and b=-18.

x=-(-18)/(2*3)

x = 3

To find the y-value, we substitute the found x into the equation.

y=3(3)^2 -18(2)

y = -9

This means the MINIMUM value is (3,-9). This is the lowest point (vertex) on the graph.