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

##### 1 Answer

Jan 22, 2017

At

#### Explanation:

Given -

#y=3-x^2-6x#

Take the first derivative

#dy/dx=-2x-6#

Find for what value of

#dy/dx=0 => -2x-6=0#

#-2x-6=0#

#-2x=6#

#x=6/(-2)=-3#

Its interpretation is when

To find the point of turn, substitute

At

#y=3-(-3)^2-6(-3)#

#y=3-9+18=12#

At

It may be minimum point or maximum point.

To find this calculate the second derivative

#(d^2y)/(dx^2)=-2 <0#

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

At

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