How to solve for a quadratic equation when given these three points...?

Since it's a quadratic equation it will be of the form

y = ax² + bx + c, and we must find a, b, and c.

Substituting the point (-1,2) into y = ax² + bx + c gives

2 = a(-1)² + b(-1) + c = 0, which can be simplified to 2 = a - b + c

Substituting the point (1,6) into y = ax² + bx + c gives

6 = a(1)² + b(1) + c = 0, which can be simplified to 6 = a + b + c

Substituting the point (2,11) into y = ax² + bx + c gives

11 = a(2)² + b(2) + c = 0, which can be simplified to 11 = 4a + 2b + c.

We have three equations and three unknowns, so we can find a, b and c using simultaneous equations.

From the first equation, c = 2 + b - a. Substitute this into 6 = a + b + c to get

6 = a + b + 2 + b - a, which simplifies to 2b = 4, and therefore b = 2.

Substitute c = 2 + b - a into 11 = 4a + 2b + c to get

11 = 4a + 2b + 2 + b - a, which simplifies to 9 = 3a + 3b, and dividing through by 3 gives 3 = a + b, and therefore a = 3 - b = 3 - 2 = 1.

Now c = 2 + b - a = 2 + 2 - 1 = 3.

We know the quadratic equation when given these points is of the form y = ax² + bx + c with a = 1, b = 2, and c = 3, so the equation is therefore y = x² + 2x + 3.