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March 26th, 2017, 05:58 AM   #1
Joined: Mar 2017
From: Mtarfa

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Quadratic Revenue Equations

I have a problem with regards to the below question. Can anybody help?

A quadratic revenue function has a graph which has a maximum value of 507 at Q = 16 and which is zero at Q = 3. Determine the second value of Q for which the revenue function is zero. Hence, find the equation for the revenue in terms of Q.

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March 26th, 2017, 06:08 AM   #2
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symmetry indicates revenue is also zero at Q = 29

$R = k(Q-3)(Q-29)$

$R(16)=507 \implies k=-3$
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March 26th, 2017, 06:11 AM   #3
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How to help you depends on what you know and what you are studying.

Is this a problem in economics or math?

Do you know calculus? What do you know about parabolas? Does the term "zero product property" ring a bell? What have you tried?

It is impossible to help if you indicate nothing about what work you have already done and what you are likely to understand.

Of course we can give answers, but if you have no idea why the answers are what they are, just an answer is not necessarily helpful for when you take your test.

Last edited by JeffM1; March 26th, 2017 at 06:14 AM.
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