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

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Karl-Olav Nyberg

Posts: 385
Registered: 12/6/04
Re: Help Needed in math
Posted: May 2, 2014 5:40 PM
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On Friday, May 2, 2014 10:36:59 PM UTC+2, quasi wrote:
> snmp wrote:
>
>
>

> >There is an easier way to look at this (even though it uses
>
> >similar principles).
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> >
>
> >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
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> >
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> >So when a = -25/12 the maxima has value 0 and so largest
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> >integer for maxima to be less than 0 is -3.
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>
>
> Yes.
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>
>
> In my first reply in this thread, I gave hints for some
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> approaches, one of which was along the lines you used above.
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>
>
> Precalculus level methods based on properties of quadratic
>
> functions make sense for this problem.
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>
>
> konyberg's solution is just an alternate method.
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>
>
> For some reason, William Elliot couldn't make sense of
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> konyberg's method, so in my reply, I tried to clarify how
>
> and why konyberg's approach works.
>
>
>
> quasi


Why is Elloit like this?



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