Question 1108467: What is the maximum or minimum value of f(x)=2x^2-8x-3 and how do you find it? Found 3 solutions by josgarithmetic, ikleyn, TeachMath:Answer by josgarithmetic(39617) (Show Source):
You can put this solution on YOUR website! recognize the form of the equation and what is should mean. Leading coefficient on x^2 term is positive, so vertex of parabola, a minimum. This occurs exactly in the middle of the two roots, regardless of complex or real. Also if you want and know how, find derivative of f(x); rate of change is 0 at this minimum point.
-------what is ?
Minimum point is (2, -11).
You can also try to factorize f(x) to find the zeros.
Use the "completing the square method:
2x^2 - 8x - 3 =
= (2x^2 - 8x + 8) - 8 - 3 =
= 2*(x - 4x + 4) - 11 =
= 2*(x-2)^2 - 11.
The minimum is achieved at x= 2. The value of the minimum is -11.
You can put this solution on YOUR website! This will have a MINIMUM value since a > 0
The minimum value occurs where x = - b/2a, or at x = - - 8/(2 * 2), or at x =8/4, or at x = 2
Substitute 2 for x in equation to get: f(2) = 2(2)^2 - 8(2) - 3
f(2) = 8 - 16 - 3
f(2), or minimum value = - 11
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