For A do I put 4 or put 4^2 (16)?

  a = -4,  b = 8,  c = 3

This is because the general template is ax^2+bx+c

Matching up the terms gets us what is shown in bold. The x's are not involved as part of a,b,c. We just focus on the numbers in front.

The graph is shown below. I used GeoGebra to make the graph.

52

Step-by-step explanation:

You can solve this by dividing the shape into three rectangles.  The horizontal rectangles have the same dimensions (11 x 2).  The vertical rectangle has the dimensions of (8 x 1).  

You can determine the dimensions by the numbers given.  11 yards as the length for the hor. rectangles are given.  2 yards as the width for the hor. rectangles are also given.  For the ver. rectangles, we can determine that the width is 1 by the bottom rectangle.  The bottom rectangle shows that the length of the parts that are not connected.  We know the total length is 11.  11 - 10 = 1.  8 yards is given.

Now use the area equation for rectangles A = lw.

For the two hor. rectangles: A = (11)(2) = 22*2 = 44 yd²

For the ver. rectangle: A = lw = (8)(1) = 8 yd²

Total area = 44 + 8 = 52 yd²