Find a quadratic function that fits the set of data points (0,−3), (2,1), (4,1). I am confused about how to go about this. Please help tonight if possible.
the general equation is ... y = ax^2 + bx + c
plug in the three points to generate three equations in a, b, and c
solve the system of equations for the three coefficients
1st one ... -3 = a(0)^2 + b(0) + c
To find a quadratic function that fits the given set of data points, we need to determine the equation of the quadratic function in the form of f(x) = ax² + bx + c.
We can start by substituting the x and y values of the data points into the equation to form a system of equations. Using the given data points (0, −3), (2, 1), and (4, 1), we get the following equations:
1st equation: f(0) = a(0)² + b(0) + c = -3 (since y = -3 when x = 0)
2nd equation: f(2) = a(2)² + b(2) + c = 1 (since y = 1 when x = 2)
3rd equation: f(4) = a(4)² + b(4) + c = 1 (since y = 1 when x = 4)
We now have a system of equations with three unknowns (a, b, and c) and three equations. To solve this system, we can use various methods such as substitution or elimination.
Let's use the method of substitution to solve the system of equations:
From the first equation: c = -3
Substituting c = -3 into the second equation:
4a + 2b - 3 = 1
4a + 2b = 4
Substituting c = -3 into the third equation:
16a + 4b - 3 = 1
16a + 4b = 4
Now we have a system of two equations with two variables. We can proceed to solve this system using any method of solving linear equations. Let's solve it using the method of substitution:
From the second equation: 2b = 4 - 4a
Substituting 2b = 4 - 4a into the first equation:
4a + (4 - 4a) = 4
4a + 4 - 4a = 4
4 = 4
The coefficients a and b have canceled out, leaving us with the equation 4 = 4, which is always true. This means that there are infinitely many solutions to this system, indicating that the data points do lie on a quadratic curve.
Now, we need to find the value of a to determine the quadratic function completely. To do this, we can choose any suitable equation and solve for a.
Let's use the second equation: 4a + 2b = 4
From the equation c = -3 (which means c = -3), we can substitute it into the third equation to get 16a + 4b = 4.
We can now solve this system of equations using elimination:
Multiply the second equation (4a + 2b = 4) by 2, giving us: 8a + 4b = 8
Subtract the new second equation (8a + 4b = 8) from the third equation (16a + 4b = 4):
(16a + 4b) - (8a + 4b) = 4 - 8
8a + 0 = -4
8a = -4
a = -4/8
a = -1/2
Now that we have found the value of a, let's substitute it back into any of the equations to solve for b. Let's use the second equation: 4a + 2b = 4
Substituting a = -1/2:
4(-1/2) + 2b = 4
-2 + 2b = 4
2b = 4 + 2
2b = 6
b = 6/2
b = 3
So we have found the values of a and b, which are a = -1/2 and b = 3. We already found earlier that c = -3.
Therefore, the quadratic function that fits the given data points is:
f(x) = (-1/2)x² + 3x - 3.