Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find the value. f(x)=-2x^2 +8x +3

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The function f(x) = -2x^2 + 8x + 3. At the minimum or maximum value of the function the first derivative with respect to x is 0.

=> f'(x) = -4x + 8 = 0

=> 4x = 8

=> x = 2

The value of x determined by solving f'(x)= 0 is a point where the value is minimum if f''(x) is positive and it is the point where the value is a maximum if f''(x) is negative.

f''(x) = -4

As this is negative at x = 2, at x = 2 the function f(x) has a maximum value equal to 11.

**f(x) = -2x^2 + 8x + 3 has a maximum value of 11.**

You may calculate the vertex of parabola of quadratic function such that:

`x = -b/(2a), y=(4ac-b^2)/(4a)`

Comparing the standard form of quadratic `ax^2+bx+c=f(x)` to the quadratic `f(x)=-2x^2 +8x +3` yields: a=-2, b=8, c=3.

Substituting -2 for a and 8 for b in `x =-b/(2a)` yields:

`x =-8/(2(-2))=8/4 =gt x = 2`

Substituting -2 for a, 8 for b and 3 for c in `y=(4ac-b^2)/(4a)` yields:

`y = (4*(-2)*3 - 64)/(4*(-2))= (-24-64)/(-8)`

`y = 88/8 = 11`

Hence, the vertex of parabola (2,11) is a critical point of function.

You need to decide if this point represents a maximum or a minimum of f(x).

Notice that the leading coefficient is negative, a=-2, hence the critical point is the maximum of the function.

**Hence, the function `f(x)=-2x^2 +8x +3` reaches its maximum at (2,11).**

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