What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36 + 13t - 60$?
\(t^2+4t-96\)
t = -b/2a
t = -4/2
t = -2
Plugging in -2 for t, we get:
\(4-8-96\)
\(-100\)
The coordinates of the vertex are (-2,-100)
The smallest value of the quadratic is -100
Answer: -100
+1234 
