# How do you find the max or minimum of f(x)=3-x^2-6x?

Jan 22, 2017

At $\left(- 3 , 12\right)$the function has a maximum.

#### Explanation:

Given -

$y = 3 - {x}^{2} - 6 x$

Take the first derivative

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x - 6$

Find for what value of x; dy/dx becomes zero

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies - 2 x - 6 = 0$

$- 2 x - 6 = 0$
$- 2 x = 6$
$x = \frac{6}{- 2} = - 3$

Its interpretation is when $x$ takes the value $0$; the slope of the curve is zero. The curve turns.

To find the point of turn, substitute $x = - 3$ in the given function

At $x = - 3$

$y = 3 - {\left(- 3\right)}^{2} - 6 \left(- 3\right)$
$y = 3 - 9 + 18 = 12$

At $\left(- 3 , 12\right)$the curve's slope is zero. The curve truns.

It may be minimum point or maximum point.

To find this calculate the second derivative

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 2 < 0$

The second derivative is less than zero. Hence the curve has a maximum.

At $\left(- 3 , 12\right)$the function has a maximum.

graph{3-x^2-6x [-32.48, 32.48, -16.24, 16.24]}