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What is the minimum or maximum of #g(x)=-x^2-6x+1#?

Answer 1

What is the min or max of #g(x) = -x^2 - 6x + 1#

Ans Max at vertex (-3, 10)

Since a < 0, the parabola opens downward, there is a Max.at vertex. x-coordinate of vertex: #x = -b/(2a) = 6/-2 = -3# y-coordinate of vertex: y = g(-3) = -9 + 18 + 1 = 10

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Answer 2

Eva Abramson

To find the minimum or maximum of ( g(x) = -x^2 - 6x + 1 ), we can use calculus. We first find the derivative of ( g(x) ), set it equal to zero, and solve for ( x ). Then, we can determine whether it corresponds to a minimum or maximum by checking the second derivative.

Find the derivative of ( g(x) ): [ g'(x) = -2x - 6 ]
Set the derivative equal to zero and solve for ( x ): [ -2x - 6 = 0 ] [ -2x = 6 ] [ x = -3 ]
Check the second derivative to determine whether ( x = -3 ) corresponds to a minimum or maximum: [ g''(x) = -2 ]

Since the second derivative is negative, ( x = -3 ) corresponds to a maximum.

So, the maximum of ( g(x) = -x^2 - 6x + 1 ) occurs at ( x = -3 ). Plugging ( x = -3 ) into ( g(x) ) gives the maximum value of the function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.