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.
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.