Inactive Tutor answered 08/25/21
2x^2 -4x - 2
take the derivative, set =0
4x -4 =0
4x =4
x =1
plug into the original equation
2(1)^2 -4(1) -2 = 2-4-2 =-4
(1,-4) is the minimum point = vertex of the upward opening parabola
Emily P.
asked 08/24/21a Determine , without graphing , whether the function has a minimum value or a maximum value. Find the minimum or maximum value and determine where it occurs .
f(x) =2x^2-4x-2
A) the function has a ___ value
B) the minimum/maximum value is___ it occurs at x=__
Inactive Tutor answered 08/25/21
2x^2 -4x - 2
take the derivative, set =0
4x -4 =0
4x =4
x =1
plug into the original equation
2(1)^2 -4(1) -2 = 2-4-2 =-4
(1,-4) is the minimum point = vertex of the upward opening parabola
Inactive Tutor answered 08/24/21
f(x) = 2x2 - 4x - 2
This is quadratic with a>0, so f(x) is facing up, so it must have a minimum
The minimum will be at the vertex
the x coordinate of the vertex = -b/2a (here a = 2, b = -4, c = -2) so you can find that
then substitute this x value into f(x) to find the minimum value
[ you can check your answers by graphing on Desmos ]
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