Question 147835
A quadratic function has a general form of 
{{{f(x)=ax^2+bx+c}}}
A quadratic function has a maximum value where its derivative equals 0.
The derivative of f(x) is,
{{{df/dx=2ax+b}}}
So now add your values that you know,
{{{f(1)=2}}}
{{{a(1)^2+b(1)+c=2}}}
1.{{{a+b+c=2}}}
and for the maximum value,
{{{2ax+b=0}}}
{{{2a(3)+b=0}}}
2.{{{6a+b=0}}}
Generally, you need three points to solve for a quadratic equation.
We can solve for the other two variables in terms of one variable.
From eq. 2,
2.{{{6a+b=0}}}
{{{b=-6a}}}
Then from eq.1,
1.{{{a+b+c=2}}}
{{{a-6a+c=2}}}
{{{c=2+5a}}}
So the general form of the solution is,
{{{f(x)=ax^2-6ax+(2+5a)}}}
There are some limits on a also.
Since the value at x=3 is a maximum, then a<0. 
{{{ graph( 300, 200, -2, 8, -20, 20, -0.5x^2+3x-.5,-x^2+6x-3, -2x^2+12x-8) }}}
Here are three graphs for a=-.5,-1,-2 just to give a sense of the family of solutions.