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 = ax^{2} + 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.
Find a quadratic model for each set of values.
Problem 1 :
(–4, 8), (–1, 5), (1, 13)
Solution :
Let the quadratic model be y = ax^{2} + 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= x^{2} + 4x + 8
Problem 2 :
(–1, 10), (2, 4), (3,–6)
Solution :
Let the quadratic model be y = ax^{2} + 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 = 10 10 - b = 10 -b = 0 b = 0 |
When a = -2 and b = 0 From (5) -2 - 0 + c = 10 c = 10 + 2 c = 12 |
So, the required quadratic function is y = -2x^{2} + 12
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM