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

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### 2 Answers

To find the minimum value of y = 4x^2 - 8x -18, we need to find the first derivative and equate it to 0. We solve for x and determine the value of the function for the value of x that we find.

y = 4x^2 - 8x - 18

y' = 8x - 8

8x - 8 = 0

=> x = 8/8

=> x = 1

Also y'' = 8 which is always positive , therefore at x = 1, we have the minimum value.

y at x = 1,

=> 4*1^2 - 8*1 - 18

=> 4 - 8 - 18

=> -22

**The minimum value of 4x^2 - 8x -18 is -22.**

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).**