What real value of t produces the smallest value of the quadratic t^2 -9t - 36?

Guest Oct 19, 2019

#1**0 **

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**

Guest Oct 19, 2019

#2**0 **

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

power27 Oct 20, 2019