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What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36 + 13t - 60$?

 Apr 9, 2024

Best Answer 

 #1
avatar+9664 
+1

Simplifying and completing the square, we have

 

\(t^2 -9t - 36 + 13t - 60\\ = t^2 + 4t - 96\\ = t^2 + 4t + 4 - 100\\ = (t + 2)^2 - 100\)

 

Note that \((t + 2)^2 \geq 0\) for any real t, and equality holds when \(t + 2 = 0\), i.e., \(t = -2\).

 

Hence, the smallest value is produced when t = -2.

 Apr 9, 2024
 #1
avatar+9664 
+1
Best Answer

Simplifying and completing the square, we have

 

\(t^2 -9t - 36 + 13t - 60\\ = t^2 + 4t - 96\\ = t^2 + 4t + 4 - 100\\ = (t + 2)^2 - 100\)

 

Note that \((t + 2)^2 \geq 0\) for any real t, and equality holds when \(t + 2 = 0\), i.e., \(t = -2\).

 

Hence, the smallest value is produced when t = -2.

MaxWong Apr 9, 2024

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