What real value of t produces the smallest value of the quadratic t^2 - 8t - 36?
To answer the question, we need to see which way the parabola is facing. Since it is facing up, the smallest point is the vertex. (if it faced down, then there would be no smallest point) Using the vertex formula (this helps us find at what value of t we get the vertex), we see that -b/2a=8/2=4.
Therefore, the smallest value of t is 4.