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).
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.