The table shows the heights y of a dropped object after x seconds. Verify that the data show a quadratic relationship. Then write a function that models the data.

Question

The table shows the heights y of a dropped object after x seconds. Verify that the data show a quadratic relationship. Then write a function that models the data.

(Table) X:| 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | Y:| 150 | 146 | 134 | 114 | 86 | 50 |

How long is the object in the air?

Answer 1

To verify this is quadratic, you must first calculate the differences between each of the table's y-values, considering the x-values increase at steady .5 intervals. So, from 150 to 146, that's -4, then from 146 to 134, -12, and so on.

The second step is calculating the differences between the differences, so from -4 to -12 that's -8, and from -12 to the subsequent number (-20), that's also -8. Since the differences between the difference for each value is the same, non-zero number (-8), the equation is quadratic. (if the difference was 0, the equation would be linear).

The second part of the problem, writing the equation, involves knowledge of the basic quadratic equation(y =ax²+bx+c). Since we have three variables, we must substitute three (x,y) values to eventually solve for all three. Using the whole x-values, we get

1. 150= a(0)²+b(0)+c --> 150 = c (FIRST VARIABLE SOLVED!)
2. 134 = a(1)² +b(1)+150 = a+b+150 --> -16=a+b
3. 86 =a(2)²+b(2)+150 = 4a+2b+150 --> -64=4a+2b --> -32=2a+b

Using the second equation, we can isolate variable a (a=-16-b), and substitute a in the third equation to get: -32=2(-16-b)+b and solve: -32=-32-2b+b --> 0 = -b = b. Substitute 0 for b in the second equation: -16=a +0, to get a=-16, b=0, and c=150. Plug these into the original equation to get y=-16x²+150. Another way to get this equation would be, instead of isolating and substituting, subtract the second two equations from each other, which will yield the same result in a quicker way (i..e (-16=a+b)-(-32=2a+b) = 16=-a --> a=-16, b=0)

Now that you have your final equation, check your work by plugging in a value not used to derive the equation, such as (.5,146). To find out how long the object is in the air, plug 0 in for the y-value and solve for x by isolating the variable:

1. 0=-16x²+150
2. -150=-16x²
3. 9.375 = x²
4. ~3.06 = x (about 3.06 seconds)