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What is the relative maximum and minimum of the function?


f(x) = 2x^2 + 28x - 8

AMinimum Value: -106 Range y > -7
BMinimum Value: -106 Range y > -106
CMinimum Value: 7

Range y > 7

DMinimum Value: -7 Range y > -7

Verified Answer :

scme1702

We equate the derivative of the function to 0 to find the x-value at its minimum:
f'(x) = 4x + 28
0 = 4x + 28
x = -7
Now, we put this value into the original equation to find the minimum value:
f(-7) = -106
The function has a positive slope so it will increase. Thus, the range will be y > -106
The answer is B