Question 622091
If this was a real life set of data, I would first plot the points to see what shape they suggest. Next, I would use some computer software to do linear regression and find the fit of the functions that I suspect best fit the data points.
If I only had pencil and paper. I would hand-draw a smooth curve that fit the points, and use my knowledge of math to figure out an equation for the function that represents the data.
The problem says it's a quadratic function, so we need to do none of that.
 
1. Three points determine a quadratic function (we may have to solve a system of equations to find coefficients), and all points would be used to verify the fit.
A quadratic function can be represented as
{{{y=ax^2+bx+c}}} or
{{{y=a(x-k)2+h}}} with {{{x=k}}} being the axis of symmetry and (k,h) being the coordinates of the vertex.
If y=0 for x=s and x=t, then quadratic polynomial can be factored, and the function could be represented as
{{{y=a(x-s)(x-t)}}}.
We have that last option, and it may be the easiest way to solve the problem.
We have {{{y=a(x-11)(x-5)}}} and we just have to find a.
For x=0,
{{{y=a(-11)(-5)=55a=55}}} --> {{{a=1}}}
So {{{y=(x-11)(x-5)}}} or {{{y=x^2-16x+55}}} could be the answer.
We may want to express it in another way,
completing the square for {{{x^2-16x}}} like this
{{{y-55=x^2-16x}}} -->  {{{y-55+64=x^2-16x+64}}} -->  {{{y+9=(x-8)^2}}} --> {{{highlight(y=(x-8)^2-9)}}}
The second option works well too, because the data suggests that the axis of symmetry is x=8, so we expect the equation to be
{{{y=a(x-8)^2+h}}}.
With the (x,y) coordinates of a couple of points we could set up a system of equations and find a and h.
For x=9, {{{y=a(9-8)^2+h=a+h=-8}}} and
For x=11, {{{y=a(11-8)^2+h=a*3^2+h=9a+h=0}}}
give us {{{system(a+h=-8,9a+h=0)}}} which gives us a=8 and h=-9,
so we get {{{highlight(y=(x-8)^2-9)}}}
2. The y-intercept is the y value when x=0. It is y=55. It is the share value for x=0, month zero, January 2009. 
3. The x-intercepts are the x values where y=0, x=7 and x=9. The share value was zero at months 7 and 9 after January 2009. That happened on August and October 2009. 
4. The equation has a negative coefficient a, so the parabola opens up. It's concave upwards, and the graph looks like a smile. The lowest point is the vertex (8,-9). It represents the lowest share value (-0), which occurred at month 8, in September 2009. The value of a share of ACME Corporation was -$9 then.