Find the maximum value of
f(x) = -2x^2 + 12x - 14
Do i have to set this equal to o to solve?
In calculus, I don't know if you are there yet, you find the max with the derivative.
f'=0=-4x+12
or x=3
then max f() is -2(9)+12(3)-14= you do it.
Now, if you don't know about derivatives yet, then
f(x)=-2(x^2-6x +9)-14+18
=-2(x-3)^2 +4
which means the max is again 4
To find the maximum value of a quadratic function, you don't need to set it equal to zero. Instead, you can use a technique called completing the square or utilize the vertex formula. I will explain both methods so you can choose the one that you find more convenient.
Method 1: Completing the Square
1) First, rearrange the quadratic function in the form f(x) = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex and a is a constant.
f(x) = -2x^2 + 12x - 14
= -2(x^2 - 6x) - 14
= -2(x^2 - 6x + 9) - 14 + 2(9) (Adding and subtracting (b/2)^2 = (6/2)^2 = 9 to complete the square)
= -2(x - 3)^2 - 14 + 18
= -2(x - 3)^2 + 4
2) Now we can see that the vertex of the parabola is at (h, k) = (3, 4). Since the coefficient of the x^2 term is negative, the parabola opens downward, and the vertex represents the maximum value of the function. Therefore, the maximum value of f(x) is 4.
Method 2: Vertex Formula
1) The vertex formula states that the x-coordinate of the vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by x = -b/(2a).
For our function f(x) = -2x^2 + 12x - 14, the coefficient of the x^2 term is -2, and the coefficient of the x term is 12. So the x-coordinate of the vertex is x = -(12)/(2*(-2)) = 3.
2) Plug this x-coordinate value back into the function to find the y-coordinate of the vertex.
f(3) = -2(3)^2 + 12(3) - 14
= -18 + 36 - 14
= 4
Therefore, the maximum value of f(x) is 4, which occurs at x = 3.