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

Nov 20, 2016

#### Explanation:

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:

$f \left(x\right) = - 3 {x}^{2} + 12 x - 7$

The vertex form of the equation of a quadratic is:

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

where $\left(h , k\right)$ 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 $- 3 {h}^{2} + 3 {h}^{2}$:

$f \left(x\right) = - 3 {x}^{2} + 12 x - 3 {h}^{2} + 3 {h}^{2} - 7$

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

$f \left(x\right) = - 3 \left({x}^{2} - 4 x + {h}^{2}\right) + 3 {h}^{2} - 7$

Set the middle term of the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ equal to the middle term in the given equation:

$- 2 h x = - 4 x$

Solve for h:

$h = 2$

Substitute ${\left(x - h\right)}^{2}$ for the terms inside the ()s:

$f \left(x\right) = - 3 {\left(x - h\right)}^{2} + 3 {h}^{2} - 7$

Substitute 2 for every h:

$f \left(x\right) = - 3 {\left(x - 2\right)}^{2} + 3 {\left(2\right)}^{2} - 7$

Combine the constant terms:

$f \left(x\right) = - 3 {\left(x - 2\right)}^{2} + 5$

We can see that the vertex is at $\left(2 , 5\right)$. 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.

Nov 20, 2016

Max of $f \left(x\right)$ is $f \left(2\right) = 5$

#### Explanation:

First Derivative of $f \left(x\right) = - 7 - 3 {x}^{2} + 12 x$ is

$f ' \left(x\right) = - \left(2\right) \left(3 x\right) + 12$
$f ' \left(x\right) = - 6 x + 12$

Second deivative of $f \left(x\right)$ is
f"(x) = -6

At the turning point $f ' \left(x\right) = 0$

Therefore $- 6 x + 12 = 0$

$x = 2$

At the point $x = 2$
$f \left(2\right) = - 7 - 3 {\left(2\right)}^{2} + 12 \left(2\right)$
$f \left(2\right) = - 7 - 12 + 24$
$f \left(2\right) = 5$

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

Therefore , the maximum value is 5