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I don't understand how to graph quadrilateral equations.

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An algebra II method of graphing this equation is done by the following steps:

Step 1: Find the x-coordinate of the vertex and then its correspending y-coordinate

           1. x = -b/2a is the formula. The a and b apply to your polynomial

           2. In your polynomial:  a = 16   b = 190   c = 0 

           3. Hence, x = -(190)/2(16)    ----------> x = -5.9375 = -95/16  

           4. Plug this value of x in our polynomial and receive its respective y value. You get the

               vertex point. 

                     y=16(-95/16)2+190(-95/16)+0   ----------> y = -564.0625 = -9025/16

           5. Vertex (-95/16, -9025/16)

Note: The axis of symmetry (which is an imaginary line) runs vertically up and down through this point and this line is defined by x = -95/16. This means that the vertex divides the parabola in half. It also means that if you graph a point on one side of the vertex/axis of symmetry, then you will know how this same point will reflect on the other side of the vertex (keep following the steps to know more)

Step 2: Guess two x values that are to the right side of the axis of symmetry/vertex. 

            Find corresponding y values. 

             1. The axis is at x = -5.9375. Right of this would be x = 0 and x = 1

             2. If x = 0, then y =16(0)2+190(0)+0 = 0      ----> (0,0)    POINT 1

                 If x = 1, then y =16(1)2+190(1)+0 = 206  ----> (1,206) POINT 2

Step 3:
Reflect POINT 1 and POINT 2 across the axis of symmetry/vertex on the left side


            1. Whenever we reflect points from one side of the vertex to the other side, the y values will remain the same. It is only the x values of the points that will change. 

            2. The parabola is symmetrical about the axis of symmetry x= -95/16. If x = 0 (which is 95/16 units to the right of the axis of symmetry), then its reflected x value on the left side of the axis is x = -95/8 (which is 95/16 units of the axis of symmetry) Reflection of POINT 1 (-95/8, 0)

            3. Similarly, if x = 1 (which is 111/16 away from the axis), then its relfected x value on the left side of the axis is x = -103/8                Reflection of POINT 2 (-103/8, 206)    

            4. Remember that though the x values were reflected, the y values remained the same. Real Life: Let's say you were waiting at a hotel and there were 2 elevators and you take one of them and your sister takes the other. The metal pole separating the elevators could be thought of as your y axis and the ground where you waited for the elevators could be your x axis. Now notice that although you both go up to the same height (same y value), your x positions are different. Your elevator's x point could be -1 and your sister's x position could be 1. 

Step 4: 

symmetric with respect to the imaginary axis of symmetry x =-95/16

Hello Joni!

Set up a table as follows:

X            16*x^2         190*x          0            Y

  and then populate the table for a range of x's:

X            16*x^2         190*x          0            Y

0             0                   0                  0             0+0+0=0

1             16                  190              0             16+190+0=206

2             64                  380             0              64+380+0=444

3             144                570             0              144+570+0=714


You are just calculating the individual powers of each x times its coeficient at the given x in each row of the table.  Include a sufficient number of x's to disclose the behavior of the function.  This function has an x^2 so it is a parabola with a minimum value at some x.  When you are satisfied you have enough x's calculated (and find the minimum Y result) you just plot the pairs of x and y values, ie the first and last column values.

A graphing calculator does just this type of calculation.  I don't know if you are using a graphing calculator but if you do it manually as described above you will get better appreciation for the function's characteristics and an even stronger appreciation  of how slick a graphing calculator can be.

Good Luck!   BruceS