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What is the maximum value of the function y equals -6x2-12x-1?
y = -6x2 - 12x - 1We recognize this as the equation of a parabola opening downward, but we don't need to know that in order to answer the question.At the extremes of a function (local max or min), the first derivative of the function = zero.The first derivative of the given function with respect to 'x' is dy/dx = -12x -12Set -12x - 12 = 0.-x - 1 = 0x = -1y = -6x2 - 12x - 1 = -6(1) - 12(-1) - 1 = -6 + 12 - 1 = 5
Square root of 98x- square root of -147x square root of 12x?
There is no answer to this problem unless x is 0. For the suare root of 98x to be a real number, x has to be positive or zero. For the square root of -147x to be a real number, x has to be negative or zero. Seeing has x has to fit both requirements, the problem has an answer only if x is zero.
Identify the maximum value of the function y equals -6x2-12x-1?
y = - 6x2 - 12x - 1 A second degree equation graphs as a parabola, and has only one max or min. At that point, the first derivative y' = 0. dy/dx = - 12x - 12 = 0 - x - 1 = 0 ==> x = - 1 At that point, y = - 6( 1 ) - 12( - 1 ) - 1 = - 6 + 12 - 1 = 5. The max value of the function is 5, and occurs when x = -1.
What is the answer to completing the square to x2 plus 12x-11?
x2 + 12x - 11 = x2 + 12x + (12/2)2 - (12/2)2 - 11 = (x + 6)2 - 47
What number must you add to complete the square x2 12x -5?
I assume you mean,X2 + 12X - 5 = 0X2 + 12X = 5Now, halve the linear term (12), square it and add it to both sidesX2 + 12X + 36 = 5 + 36factor on the left, gather terms on the right(X + 6)2 = 41(X + 6)2 - 41 = 0==============vertex form.... (- 6, - 41)========================--------------vertexAnd that number added is 36
