What is the maximum value of the expression \(-5r^2 + 40r - 12\) for real \(r\)?
thx in advance
+6248 \(\text{probably easiest to just use the cheater formula for the vertex and evaluate the expression there}\\ \text{as it's a downward facing parabola that will be the maximum}\\ \text{the cheater formula for the vertex of $ax^2+bx+c$ is $x_v = -\dfrac b {2a}$}\\ r_v = -\dfrac{40}{2\cdot (-5)} = 4\\ f_v = -5(4^2)+40(4) - 12 = -80 +160-12 = 68\)
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