# EQUATION OF PARABOLA FROM 3 POINTS

The graphical form of quadratic function will be a parabola. To find the quadratic function passes through three given points, we have to follow the steps given below.

The quadratic equation which is in the form of

y = ax2 + bx + c

Step 1 :

Apply the given points one by one in the equation above. So, we will get three equations in three variables.

Step 2 :

By solving the system of equations in three variables, we will get the values of the variables a, b and c.

Step 3 :

Here the sign of the variable a will decide whether the parabola opens up or down.

• If a < 0, then the parabola opens down.
• If a > 0, then the parabola opens up.

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

 When a = -2 From (5)-5(-2) - b = 1010 - b = 10-b = 0b = 0 When a = -2  and b = 0From (5)-2 - 0 + c = 10c = 10 + 2c = 12

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

## Recent Articles

1. ### Finding Range of Values Inequality Problems

May 21, 24 08:51 PM

Finding Range of Values Inequality Problems

2. ### Solving Two Step Inequality Word Problems

May 21, 24 08:51 AM

Solving Two Step Inequality Word Problems