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Topic: Help Needed in math
Replies: 25   Last Post: May 2, 2014 6:01 PM

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Posts: 12,067
Registered: 7/15/05
Re: Help Needed in math
Posted: May 2, 2014 4:36 PM
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snmp wrote:

>There is an easier way to look at this (even though it uses
>similar principles).
>Axis of symmetry for ax^2 + bx +c = 0 is x = -b/2a = 5/2a

You need to consider the special case a = 0.

>Value of function at axis symmetry (gives either maxima when
>a < 0 (curve is concave down) and minima when a > 0 (curve is
>concave up)): a(5/2a)^2 -5(5/2a) - 3 = 0 --> a = -25/12
>So when a = -25/12 the maxima has value 0 and so largest
>integer for maxima to be less than 0 is -3.


In my first reply in this thread, I gave hints for some
approaches, one of which was along the lines you used above.

Precalculus level methods based on properties of quadratic
functions make sense for this problem.

konyberg's solution is just an alternate method.

For some reason, William Elliot couldn't make sense of
konyberg's method, so in my reply, I tried to clarify how
and why konyberg's approach works.


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