+0

Here is a quick question. thx

0
45
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What is the maximum value of the expression $$-5r^2 + 40r - 12$$ for real $$r$$?

Aug 17, 2019

#1
+5766
+1

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

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Aug 17, 2019
#2
+1

Thank you SO MUCH!

Aug 17, 2019