What real value of t produces the smallest value of the quadratic t^2 -12t - 36?
+2407 t^2 -12t - 36 = (t-6)^2 - 72
Do you have an idea on how to solve this now?
=^._.^=
+876 This problem can be rephrased into: What is the x-value of the vertex of the parabola $t^2 - 12t - 36$?
The formula for $x$ is $- \frac{b}{2a},$ for polynomial $ax^2 + bx + c.$
We have: $-\frac{-12}{2} = \boxed{6}$
You can see that the graph shown here: https://www.desmos.com/calculator/fztszalids is least when $x=6.$
