Question 314772
The parabola is already in vertex form,
{{{y=a(x-h)^2+k}}}
where (h,k) is the vertex. 
It's just a bit hidden.
{{{y=-3x^2+6}}}
{{{y=-3(x-0)^2+6}}}
The vertex is (0,6).
{{{ graph( 300, 300, -5, 5, -5, 10, -3x^2+6) }}}
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 x-3/x+5<0.
Break up the number line into 3 regions.
Region 1: ({{{-infinity}}},{{{-5}}})
Region 2: ({{{-5}}},{{{3}}})
Region 3: ({{{3}}},{{{infinity}}})
For each region, choose a point (not an endpoint) and test the inequality.
If the inequality is solved, that region is part of the solution.
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Region 1: {{{x=-10}}}
{{{(x-3)/(x+5)<0}}}
{{{(-10-3)/(-10+5)<0}}}
{{{(-13)/(-5)<0}}}
{{{13/5<0}}}
False, this region is not part of the solution.
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Region 2: {{{x=0}}}
{{{(x-3)/(x+5)<0}}}
{{{(0-3)/(0+5)<0}}}
{{{-3/5<0}}}
True, this region is part of the solution.
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Region 3: {{{x=5}}}
{{{(x-3)/(x+5)<0}}}
{{{(5-3)/(5+5)<0}}}
{{{2/10<0}}}
False, this region is not part of the solution.
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Only Region 2 is the solution region.
({{{-5}}},{{{3}}})
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{{{ drawing( 300, 300, -10, 10, -5, 5,grid(1),blue(line(-5,10,-5,-10)),blue(line(3,10,3,-10)), graph( 300, 300, -10, 10, -5,5, (x-3)/(x+5)
 ) )}}} 
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The curve is below the x-axis(< 0) on in the region (-5,3).