Which quadratic function best fits this data?

x y

1 250

2 289

3 316

4 335

5 320

6 290

y=−9.16x2+73.04x+183.3

y=9.16x2−73.04x+183.3

y=9.16x2+73.04x+183.3

y=−9.16x2−73.04x+183.3

Step-by-step explanation:

As we move from left to right, and as x accordingly varies from 1 to 6 , we see the y value first increasing, then reaching an actual or approximate maximum of 335 for x = 4, and then decreasing as x continues to increase.  

This is the graph of a parabola opening down, which focuses our attention on answer choices A and D only.

Let's check out A:  Use synthetic division to determine the approximate value of y when x = 3 (this is a somewhat arbitrary choice for x):

   3    /   -9.16    73.04    183.3

                         -27.48    -136.68

             -9.16     -45.56    46.62    Thus, the value of the first function at

                                                        x = 3 is approx. 46.62, which is rather

                                                        far from the expected 316.

Let's check out D:    y = -9.16x^2 - 73.04x + 183.3

Again let x = 3.  The smaller the remainder in synthetic division, the closer this equation predicts the y value:

     3    /     -9.16       -73.04         183.3

                                 -27.48        -301.56

                  -9.16       -100.52        -118.26    This remainder is even further

                                                                      from the expected value (316)

The first equation is a (somewhat) better fit than is the fourth.  The second and third equations are totally inappropriate because their graphs open UP instead of down.

y = −9.16x2 + 73.04x + 183.3

This is the correct answer! I just took the test! Good luck!

y=mx+b

step-by-step explanation:

m slope, b y-intercept