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

 Oct 19, 2019
 #1
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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

 Oct 19, 2019
 #2
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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

 Oct 20, 2019

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