What is the maximum point of y = 3x^2 - 18x
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`
`y = 3x^2 - 18x`
Applying the partial differentiation with respect to x
`( del y )/(del x) = 6*x - 18`
As we need to get the max point we need to apply ( del y )/(del x) = 0
=> `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 = 3
To find the y-value, we substitute the found x into the equation.
y = -9
This means the MINIMUM value is (3,-9). This is the lowest point (vertex) on the graph.