Formulate the quadratic function that contains the points (-1,2), (0,-1) and (2,5).

The idea here is to take all three of those points and sub them into the general form of the quadratic equation and solve 3 equations by doing them 2 at a time.  The general form is y = ax^2 + bx + c. If we set up the first two equations together using the x and y coordinates of the first 2 coordinates given, the system looks like this: 2 = a(-1)^2 + b(-1) + c
                                   -1 = a(0)^2 +b(0) + c
Solving the second one gives you that c = -1, so we already know c. Solving the first equation we get 2 = a - b - 1,  or 3 = a - b, subbing in a -1 for c. The next equation, using the last coordinate, looks like this: 5 = a(2)^2 + b(2) - 1 and
5 = 4a + 2b - 1 and 6 = 4a + 2b. Now we have these two equations left:
3 = a - b and 6 = 4a + 2b. Solve the first equation for b to get it in terms of a only: b = a - 3. Now sub in that b value for b in the second equation:
6 = 4a + 2(a - 3) and 6 = 6a - 6 and 12 = 6a so a = 2. Now we have a and c. Sub in the a value of 2 into b = a - 3 to get 3 = 2 - b. Solve for b to get that b = -1. So our equation in the end finally is y = 2x^2 - x - 1.

r   = 2.65m

good luck!

cutie~