Question

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

Answer 1

Step-by-step explanation:

To find extreme values, take the first derivative

$y = - 3 {x}^{2} + 16x - 16$

Solve the equation for x.

$x = \frac{ - 16 + \sqrt{256 - 192} }{ - 6}$

or

$x = \frac{ - 16 - \sqrt{256 - 192} }{ - 6}$

$x = \frac{ - 16 + \sqrt{64} }{ - 6}$

or

$x = \frac{ - 16 - \sqrt{64} }{ - 6}$

We then get

$x = \frac{4}{3}$

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.