Question 143395
First thing to notice is that two of the points are on the x-axis and that tells us that we have a parabola whose axis is perpendicular to the x-axis, i.e. vertical, and that the x-coordinates of these two points are the zeros of the desired polynomial function.


Knowing that we have two zeros for a 2nd degree polynomial, we can derive A quadratic function simply by multiplying {{{(x+3)(x-4)=x^2-x-12}}}.  However, this misses the mark because if you calculate the y-intercept you get (0,-12) instead of the desired (0,2).


Fortunately, you can multiply any polynomial by any constant and not change the zeros.  {{{x^2+4x-5}}}, {{{2x^2+8x-10}}}, and {{{x^2/4+x-5/4}}} all have the same zeros, namely 1 and -5.


So for the function in question, we need to answer, "What can we multiply by so that the constant term will be 2?"  Answer: {{{-1/6}}}.


So:  {{{f(x)=-x^2/6+x/6+2}}} is the desired function.


Check:
{{{f(0)=-(0)^2/6+0/6+2=2}}} 


{{{f(-3)=-(-3)^2/6+(-3)/6+2=-9/6-3/6+2=0}}}


{{{f(4)=-(4)^2/6+(4)/6+2=-16/6+4/6+2=0}}}