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

 Jul 26, 2019
 #1
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To find the minimum value, you can complete the square.

 

Using the formula \(a(t+d)^2+e\) where \(d=\frac{-b}{2a}\) and \(e=c-\frac{b^2}{4a}\), the completed square would be \((t-\frac{9}{2})^2-45\). Since squares cannot be negative, we see that the minimum value would be where \((t-\frac{9}{2})^2\) is 0. Thus \(t=\frac{9}{2}\)

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 Jul 26, 2019
 #2
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There are a couple of other ways....   this is a bowl shaped parabola (because the leading coefficient is positive)

 

- b / 2a

= 9/2

 

 

 

or take the derivative and = 0

2t-9 = 0

2t=9

t=9/2

 Jul 26, 2019

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