Question 828362
Some of your steps are not clear.  You could use the given three points to form three equations with unknown coefficients and use very simple linear algebra skills to find the values of these coefficients.  


You can also use the quality of symmetry for a quadratic equation of a parabola, and find the x value of the center, so you know the vertex is on that line of symmetry.  Having points (1,-2) and (2,-2), you know that because of symmetry, this parabola has a vertex on {{{x=3/2}}}.  


Looking at the other given point on the parabola, (3,-4), you understand that the "a" coefficient on x^2 is negative; the value of the function decreases as x moves away from {{{x=3/2}}}, and the function has a MAXIMUM at {{{x=3/2}}}.


LOOK AT the standard form equation for your so-far-unfinished function.
{{{y=a(x-3/2)^2+k}}}.
Plug in the point (3,-4).
{{{-4=a(3-3/2)^2+k}}}, and simplify.
{{{-4=a(9/4)+k}}}
{{{-16=9a+4k}}}
{{{9a+4k=-16}}}-----nice linear equation.
NOW, use that same standard form equation and substitute for either (1,-2) OR (2,-2).  No matter which; just choose either.
{{{-2=a(2-3/2)^2+k}}}
{{{-2=a(1/4)+k}}}
{{{-8=a+4k}}}
{{{a+4k=-8}}}----another very nice linear equation.
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Solve this system of two linear equations in the unknowns, a and k.
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9a+4k=-16
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a+4k=-8
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Once solved for a and k, finish your standard form equation of {{{y=a(x-3/2)^2+k}}}




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note:  {{{a=-1}}} and therefore {{{k=-7/4}}}, easily found through Elimination Method.  Finished standard form parabola, {{{highlight(y=-(x-3/2)^2-7/4)}}}.