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If $t$ is a real number, what is the maximum possible value of the expression $-t^2 + 8t -4 +5t^2 - 4t + 18$?

 Aug 11, 2024
 #1
avatar+1858 
+1

The maximum possible value of the expression occurs at the vertex of the graph. 

So let's find the y value of the vertex.

First, simplifying and combining all like terms, we get the equation

\(4t^{2}+4t+14\)

 

Now, the x value of the vertex of the graph is \(\frac{-b}{2a}\), so we have

\(\frac{-4}{8} = -1/2\)

 

Plugging this value back into equation and subbing out t, we get

\(y = 13\)

 

So 13 is the our answer. 

 

Thanks! :)

 Aug 12, 2024
edited by NotThatSmart  Aug 12, 2024
 #2
avatar+1858 
+1

Sorry, just realized, for this question, there IS no max value. 

The MIN value is 13 since the parabola open upwards. 

 

My bad.

NotThatSmart  Aug 12, 2024
edited by NotThatSmart  Aug 12, 2024

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