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

justaguide | College Teacher | (Level 2) Distinguished Educator

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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)

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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)

neela | High School Teacher | (Level 3) Valedictorian

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To verify if the quadratic has extreme values

y=-3x^2+2x+5.

Let y(x) = -3x^2+2x+5.

y(x) = has the extreme values when y'(x) = 0.

=>  (-3x^2+2x+5)' = 0

=> (-6x+2) = 0.

=> 6x= 2.

6x/6 = 2/6 = 1/3.

Therefore  f(1/3) = -3(1/3)^2+2(1/3)+ 5 = -3/9+2(1/3)+5 = 16/3.is an extreme value of f(x).

f"(x) = -6 for all x, So f'(1/3) = -16/3 < 0.

So f(1/3) = 16/3 is the maximum value of f(x). Or for x= 1/3 , f(x) attains its maximum.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The extreme value of the quadratic is represented by the vertex of the parabola, that is the graph of the quadratic function.

We'll re-write the equation of the function:

y = f(x)

f(x) = -3x^2 + 2x + 5

We notice that the coefficient of x^2 is a = -3, so the vertex is a maximum point.

We'll calculate the coordinates of the vertex using the formula:

xV = -b/2a

yV = -delta/4a

a = -3, b = 2 , c = 5

xV = -2/-6

xV = 1/3

yV = -(b^2 - 4ac)/4a

yV = (-60 - 4)/-12

yV = -64/-12

yV = 16/3

The extreme value of the given function is the maximum point of the prabola and it has the coordinates (1/3 ; 16/3).