need help on number 5 thanks?

a) So put the numbers for years on x and the rates on y.

b) If vertex is at (h, k), then the equation is f(x) = a(x-h)^2 + k
f(x) = a(x-4)^2 + 4.4

We use the 2nd point (9, 3.1) to find a
3.1 = a(9-4)^2 + 4.4
a = (3.1 - 4.4) / 5^2 =
This will be a very poor estimate because the speed at which dropout rate is changing should increase further from the vertex in a quadratic function. Here it decreases rapidly. A calculator will find a better one (not that a quadratic functiona should be used in the first place)

This is best done on a graphing calculator or spreadsheet.

(a) See the source link
(b) Using the given points, we have
.. f(x) = a(x -4)^2 +4.4 . . . . . must fit (9, 3.1)
.. 3.1 = a(9 -4)^2 +4.4
.. (3.1 -4.4)/25 = a = -0.052
The function is
.. f(x) = -0.052(x -4)^2 +4.4

(c) My TI-83/84 gives the constants for f(x) = ax^2 +bx +c as
.. a = -0.0023809524
.. b = -0.1357142857
.. c = 4.480952381
All are plotted at the source link. The calculator function seems a better fit.