What is the minimum value of 4x^2-8x-18 ?

To find out the extreme value of a function, we'll have to
calculate the first derivative of the function.

Let's find
the first derivative of the function
f(x):

f'(x)=( 4x^2-8x-18)'=(4x^2)'-(8x)'-(18)'

f'(x)=8x-8

Now we
have to calculate the equation of the first
derivative:

8x-8=0

We'll
divide by 8:

x-1 =
0

x=1

That means that the
function has an extreme point, for the critical value
x=1.

f(1) = 4*1^2 - 8*1 -
18

f(1) = 4 - 8 -
18

We'll combine like
terms:

f(1) =
-22

The extreme point of the function is a
minimum point whose coordinates are: (1 ; -22).