In an equation of the form with k=ax^2+bx+c with a>0, the least possible value of k occurs at x=-b/(2a). In the equation k=(6x+12)(x-8), what is the least possible value for k?
+701 the vertex = (h,k)
h= -b/2a
you want to simplify your expresion (6x+12)(x-8)
and when you do so you get something that is x^2+bx+c form (standard) so you know that if you were to graph this parabola, it would have a minimum since your a value is positive
f(x)+ a(x-h)^2+k
you want to then take your standard form equation and you can either put it into vertex form, or you can use -b/2a to get your h value which is the x value of your vertex, and then take the f(-b/2a) meaning take that value you got from your -b/2a and plug it into your equation as x and solve and then you get your k value.
OR
you can just write it into vertex form and you can then extract (h,k) and done!!
hope this helped you somehow :)
