Question 887363
{{{ax^2+bx+c=y}}}


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{{{0+0+c=3}}}
{{{a*1+b*1+c=4}}}
{{{a(-1)^2+b(-1)+c=-6}}}
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{{{0+0+c=3}}}
{{{a+b+c=4}}}
{{{a-b+c=-6}}}
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The variables in this now  become a, b, c.
The y-intercept point and its equation allow eliminating c so you can form two equations in just the unknowns a and b.


You already see {{{highlight(c=3)}}}.


{{{a+b=4-c=4-3}}}, {{{highlight_green(a+b=1)}}}.


{{{a-b=-6-c=-6-3}}}, {{{highlight_green(a-b=-9)}}}.


Add those two and you can solve for a.
{{{highlight(a=-5)}}}.


Use either of the a & b equations to find b.
{{{a+b=1}}}
{{{b=1-a}}}
{{{b=1-(-5)}}}
{{{highlight(b=6)}}}.


General equation now, {{{highlight(y=-5x^2+6x+3)}}}.
Is that factorable?
Discriminant, {{{6*6-4*(-5)*3=36+60=96}}}, quadratic is NOT factorable.


You still want the vertex maximum point in order to be able to graph y.
You have your choice of completing the square to convert into standard form; or you can use the general solution of a quadratic equation to find the zeros and the vertex will be in the exact middle of the two solutions.


This may help:  <a href="http://www.algebra.com/tutors/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev">Completing the Square to Solve General Quadratic Equation</a>