EQUATION OF PARABOLA FROM 3 POINTS

Find a quadratic model for each set of values.

Problem 1 :

(–4, 8), (–1, 5), (1, 13)

Solution :

Let the quadratic model be y = ax² + bx + c

The parabola passes through the given points, so we can apply the points one by one.

It passes through (-4, 8).

8 = a(-4)² + b(-4) + c

8 = 16a - 4b + c

16a - 4b + c = 8 -----------(1)

It passes through (-1, 5).

5 = a(-1)² + b(-1) + c

5 = a - b + c

a - b + c = 5 -----------(2)

It passes through (1, 13).

13 = a(1)² + b(1) + c

13 = a + b + c

a + b + c = 13 -----------(3)

Subtracting (1) and (2), we get

16a-4b-a+b = 8-5  

15a - 3b = 3 -----(4)

Subtracting (2) and (3), we get

a - b + c - a - b - c = 5 - 13

-2b = -8

b = 4

By applying the value of b in (4), we get

15a - 3(4) = 3

15a - 12 = 3

15a = 15

a = 15/15

a = 1

Applying a = 1 and b = 4 in (2)

1 - 4 + c = 5

-3 + c = 5

c = 8

Applying the values of a, b and c in the general form of quadratic function, we will get

y= x² + 4x + 8

Problem 2 :

(–1, 10), (2, 4), (3,–6)

Solution :

Let the quadratic model be y = ax² + bx + c

The parabola passes through the given points, so we can apply the points one by one.

It passes through (-1, 10).

10 = a(-1)² + b(-1) + c

10 = a - b + c

a - b + c = 10 -----------(1)

It passes through (2, 4).

4 = a(2)² + b(2) + c

4 = 4a + 2b + c

4a + 2b + c = 4 -----------(2)

It passes through (3, -6).

-6 = a(3)² + b(3) + c

-6 = 9a + 3b + c

9a + 3b + c = -6 -----------(3)

Subtracting (1) and (2), we get

a - b - 4a - 2b = 10 - 4

-3a - 3b = 6

-a - b = 2 -----(4)

Subtracting (2) and (3), we get

4a + 2b - 9a - 3b = 4+6

-5a - b = 10 -----(5)

Subtracting (4) and (5)

-a - b + 5a + b = 2 - 10

4a = -8

a = -2

So, the required quadratic function is y = -2x² + 12