Question 191813
Given:
{{{V(x) = x^2-6x+13}}} Find the minimum value of x.
The minimum value of x will occur at the vertex of the upward-opening parabola.
This is found by:
{{{x = (-b)/2a}}} where a = 1 and b = -6, so...
{{{x = -(-6)/2(1)}}}
{{{highlight(x = 3)}}}
The lowest value of V(x) is found by substituting the minimum value of x (x = 3) into the given function {{{V(x) = x^2-6x+13}}} and solving for V.
{{{V(x) = x^2-6x+13}}} Substitute x = 3.
{{{V(3) = (3)^2-6(3)+13}}} Simplify.
{{{V(3) = 9-18+13}}}
{{{V(3) = 22-18}}}
{{{V(3) = 4}}}
The lowest value of V(x) is 4.
Since x = the number of months after January, 2004, the lowest value of V(x) will be January + 3 months or April of 2004.