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What real value of t produces the smallest value of the quadratic t^2 -12t - 36?

 Jun 23, 2021
 #1
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t^2 -12t - 36 = (t-6)^2 - 72

Do you have an idea on how to solve this now?

 

=^._.^=

 Jun 23, 2021
 #2
avatar+736 
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This problem can be rephrased into: What is the x-value of the vertex of the parabola $t^2 - 12t - 36$?

The formula for $x$ is $- \frac{b}{2a},$ for polynomial $ax^2 + bx + c.$

We have: $-\frac{-12}{2} = \boxed{6}$

 

You can see that the graph shown here: https://www.desmos.com/calculator/fztszalids is least when $x=6.$

 Jun 23, 2021

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