How do you find the max or minimum of `f(x)=-7-3x^2+12x`?

Because this question is in precalculus, I am going to assume that you have not, yet, studied differential calculus. Therefore, I am going to show you how to find the minimum or maximum of a quadratic by finding the vertex:

Let's begin by writing the quadratic equation in standard form:

`f(x) = -3x^2 + 12x - 7`

The vertex form of the equation of a quadratic is:

`f(x) = a(x - h)^2 + k`

where `(h, k)` is the vertex and "a" is the coefficient of the `x^2` term in standard form.

Because "a" in -3, we shall add 0 to the original equation in the form `-3h^2 + 3h^2`:

`f(x) = -3x^2 + 12x -3h^2 + 3h^2 - 7`

This allows us to factor a -3 from the first 3 terms:

`f(x) = -3(x^2 - 4x + h^2) + 3h^2 - 7`

Set the middle term of the right side of the pattern `(x - h)^2 = x^2 - 2hx + h^2` equal to the middle term in the given equation:

`-2hx = -4x`

Solve for h:

`h = 2`

Substitute `(x - h)^2` for the terms inside the ()s:

`f(x) = -3(x - h)^2 + 3h^2 - 7`

Substitute 2 for every h:

`f(x) = -3(x - 2)^2 + 3(2)^2 - 7`

Combine the constant terms:

`f(x) = -3(x - 2)^2 + 5`

We can see that the vertex is at `(2, 5)`. Because "a" is negative we know that this is a maximum. If are were positive, then the vertex would be a minimum.

The maximum value that this quadratic achieves is 5, at the x coordinate 2.

First Derivative of `f(x) = -7 - 3x^2 + 12x` is

`f '(x) = - ( 2 )(3x) +12`
`f'(x) = - 6x + 12`

Second deivative of `f(x)` is
`f"(x) = -6`

At the turning point `f'(x) = 0`

Therefore `- 6 x + 12 = 0`

`x = 2`

At the point `x = 2`
`f (2) = -7 - 3(2)^2 + 12 (2)`
`f(2) = -7 - 12 + 24`
`f(2) = 5`

`f"(2) = - 6` imply the point (2, 5) is a maximum#

Therefore , the maximum value is 5