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

Expert Answers

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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.

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