Question 11856
Let x = first number
x+6 = second number


Let y = product of the numbers
y = x(x+6)
{{{y = x^2 + 6x}}}


Graph the equation of this parabola, and find the lowest value of y, which is the vertex of the parabola.  Algebraically, the vertex will always be at {{{x = -b/(2a)}}} (where a=1, b=6 as in the quadratic formula!) so {{{x = -6/2 = -3}}}.  Also halfway between the x intercepts, which would be at x=0 and x= -6.  The minimum value of the product of the numbers would be 
{{{y = (-3)^2 + 6(-3) = 9-18 = -9}}}.


{{{graph (300, 300, -10, 10, -10, 10, x^2 + 6x)}}}


R^2 at SCC