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What is the maximum value of the expression \(-5r^2 + 40r - 12\) for real \(r\)?

 

thx in advance

 Aug 17, 2019
 #1
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\(\text{probably easiest to just use the cheater formula for the vertex and evaluate the expression there}\\ \text{as it's a downward facing parabola that will be the maximum}\\ \text{the cheater formula for the vertex of $ax^2+bx+c$ is $x_v = -\dfrac b {2a}$}\\ r_v = -\dfrac{40}{2\cdot (-5)} = 4\\ f_v = -5(4^2)+40(4) - 12 = -80 +160-12 = 68\)

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 Aug 17, 2019
 #2
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Thank you SO MUCH!

 Aug 17, 2019

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