Question 1069344
{{{y=a(x+2)(x-3)(x-2/5)}}}
So when {{{x=0}}}, {{{y=6}}},
{{{6=a(0+2)(0-3)(0-2/5)}}}
{{{6=a(12/5)}}}
{{{a=5/2}}}
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{{{y=(5/2)(x+2)(x-3)(x-2/5)}}}
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or expanding,
{{{y=(5/2)x^3-(7/2)x^2-14x+6}}}