What real value of t produces the smallest value of the quadratic t^2 -9t - 36?
Solve for t:
t^2 - 9 t - 36 = 0
The left hand side factors into a product with two terms:
(t - 12) (t + 3) = 0
Split into two equations:
t - 12 = 0 or t + 3 = 0
Add 12 to both sides:
t = 12 or t + 3 = 0
Subtract 3 from both sides:
t = 12 or t = -3
+247 To find the smallest value, you need to find the vertex of the porabola. The vertex is the point that is the lowest on the porabola (or highest if a is negative). The x value produces the lowest (or highest) value of the porabola, so we only need to find one half of the point.
To find the x value, plug the coefficients into \({-b \over 2a}\)
\({-(-9) \over 2(1)}\)
=9/2
=4.5 will produce the lowest value of the porabola
