first off, let's find the zeros, keeping in mind that, the vertex of the parabola, will be half-way between the zeros.
[tex]\bf y=3x^2+x-2\implies 0=3x^2+x-2\implies 0=(3x-2)(x+1)
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0=3x-2\implies 2=3x\implies \boxed{\cfrac{2}{3}=x}
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and
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0=x+1\implies \boxed{-1=x}[/tex]
now, the zeros are there, now, what's half-way? or the midpoint of that interval? well, from -1 to 0, is say 3/3, from 0 to 2/3 is 2/3, we add both to get 3/3 + 2/3 the interval is 5/3.
now, what's half of 5/3? [tex]\bf \cfrac{\quad\frac{5}{3} \quad }{2}\implies \cfrac{\quad\frac{5}{3} \quad }{\frac{2}{1}}\implies \cfrac{5}{3}\cdot \cfrac{1}{2}\implies \cfrac{5}{6}[/tex]
now, 5/3 is the same as 10/6, so if we move from say the -1 spot over 5/6 towards the 0, we'll end up at -1/6.
therefore half of 5/3 is 5/6, and if we move from either "zero" to the middle by 5/6 units, we'll end up at -1/6, check the picture below.
therefore, the vertex is right there, above -1/6, when x = -1/6, what is that anyway?
[tex]\bf y=3x^2+x-2\qquad \boxed{x=-\frac{1}{6}}\qquad y=3\left( -\frac{1}{6} \right)^2+\left( -\frac{1}{6} \right)-2
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y=3\left( -\frac{1}{6} \right)\left( -\frac{1}{6} \right)-\frac{1}{6}+2\implies y=3\left( \frac{1}{36} \right)-\frac{1}{6}-2
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y=\cfrac{1}{12}-\cfrac{1}{6}-2\impliedby \textit{our LCD is 12}\implies y=\cfrac{1-2-24}{12}
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\boxed{y=\cfrac{-25}{12}}[/tex]
