Where does the extreme value of y = -3x^2 + 2x + 5 lie?

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To find the extreme value of y = -3x^2 + 2x + 5, we have to find the derivative y'.

y' = -3*2x + 2

Equating this to zero,

-6x + 2 = 0

=> x = 2/6 = 1/3

When x = 1/3, y = -3*(1/3)^2 + 2/3 + 5

= -3 / 9 + 2/3 + 5 = -1/3 + 2/3 + 5 = 1/3 + 5

= 16/3

Therefore the extreme point is (1/3 , 16/3)

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Given the function:

y= -3x^2 + 2x + 5

To find the extreme values, first we will find the derivative y'.

==> y' = -6x + 2

Now we will find the critical values which is the derivatives zeros.

==> -6x + 2 = 0

==> -6x = -2 ==> x= 2/6 = 1/3

Then, the function has extreme value when x= 1/3

==> y(1/3) = -3(1/3)^2 + 2(1/3) +5 = -1/3 + 2/3 + 15/3 = 16/3

We notice that the sign of x^2 is negative. Then, the function has a maximum values at y(1/3) = 16/3.

Then,the maximum values is the point (1/3, 16/3)

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