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# someone help?

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If \(t\) is a real number, what is the maximum possible value of the expression \(-t^2 + 8t -4\)?

Jul 28, 2019

#1
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We have the form  at^2 + bt + c   where   a  = -1   and b  =  8

This is a parabola that turns downward.....the y value of the vertex will be the max

The "t"  value of the vertex =   -b / [ 2a ]  =  - [ 8 ] / [2 (-1) ]  = -8/ -2  = 4

So......the max value is    -(4)^2 + 8(4) - 4   =   -16 + 32 - 4  =   -16 + 28  =  12

The vertex = ( t, y)  = (4, 12)

Here's the graph showing this : https://www.desmos.com/calculator/6grvgwmlqj   Jul 28, 2019
#2
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Thanks CPhill!

Jul 28, 2019
#3
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We can also complete the square. We can factor out a -1 from the given expression. We have -(t^2-8t+4).  We can work inside the parenthesis. We can try to make the quadratic a perfect square. We have: t^2-8t+4=(t-4)^2-16+4=(t-4)^2-12. Remember the negative ;) We now have: -(t^2-4)^2+12. Since the negative of a positive number (a square) is always negative, the maximum it can be is 0. Therefore, the only possible maximum is 12. Hope this gives helpful insight :)

Jul 28, 2019