maximum point of a curveFind the maximum point of the curve y = x^2 - 8x + 16
You may evaluate the maximum of the given function using derivative of the function, such that:
`f'(x) = 0`
Evaluating `f'(x) ` yields:
`f'(x) = (x^2 - 8x + 16)' => f'(x) = 2x - 8`
`f'(x) = 0 => 2x - 8 = 0 => 2x = 8 => x = 4`
Hence, the function `f(x) = x^2 - 8x + 16` reaches its maximum at `x = 4.`
The maximum point of these quadratic function is represented by the vertex of the function.
The graph of the quadratic function is a parabola and the coordinates of the parabola vertex are:
V(-b/2a;-delta/4a), where a,b,c are the coefficients of the function and delta=b^2 -4*a*c.
y=f(x)=x^2 - 8x + 16
We'll identify the coefficients:
a=1, 2a=2, 4a=4
delta =64 - 64
delta = 0
We notice that the x coordinate is positive and y coordinate is 0, so the vertex of parabola is located on the right side of x axis: V(4;0).