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Determine the zeros, sign and extreme values of the function y = -x³ + 8x²-16x.​

User SiKing
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1 Answer

11 votes

Explanation:

To find extreme values, take the first derivative


y = - 3 {x}^(2) + 16x - 16

Solve the equation for x.


x = ( - 16 + √(256 - 192) )/( - 6)

or


x = ( - 16 - √(256 - 192) )/( - 6)


x = ( - 16 + √(64) )/( - 6)

or


x = ( - 16 - √(64) )/( - 6)

We then get


x = (4)/(3)

or


x = 4

So our extreme values occur at x=4/3 and x=4

Zeroes:


- {x}^(3) + 8 {x}^(2) - 16x


- x( {x}^(2) - 8x + 16)


- x(x - 4) {}^(2)


- x = 0


(x - 4) {}^(2) = 0


x = 4

So we have a double root at 4, and a root of 0. We have two positive roots and one non negative root

End behavior:

Since we have a negative leading coefficient and a odd degree, as x becomes more positive, y will become negative and as x becomes more negative, y will become positive.

User Kiwiron
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7.1k points