Question 127491
Find the maximum/minimum point (vertex) of:
{{{f(x) = x^2-16x+66}}}
First, notice that the coefficient of the {{{x^2}}} term is +1, so this tells you that the parabola opens upwards, therefor, the vertex will be at the minimum point of the curve.
The x-coordinate of the parabola is given by: {{{x = -b/2a}}}, so...
{{{x = -(-16)/2(1)}}}
{{{x = 16/2}}}
{{{x = 8}}} Substitute this into the original equation to find the y-coordintae of the vertex:
{{{y = 8^2-16(8)+66}}}
{{{y = 64-128+66}}}
{{{y = 130-128}}}
{{{y = 2}}}
The location of the vertex is at (8, 2) and is a minimum.