SOLUTION: find a quadratic function that fits the set of data points (2,0) (4,3) (12,-5)

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Question 619751: find a quadratic function that fits the set of data points (2,0) (4,3) (12,-5)
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

find a quadratic function that fits the set of data points (2,0) (4,3) (12,-5)

 y = ax^2 + bx + c = 

  4a + 2b +  c = 0

 16a + 4b +  c = 3

144a + 12b + c = -5  (-1/4, 3, -5)

y = (-1/4)x^2 + 3x - 5




 
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables




  

  

  system%284%2Ax%2B2%2Ay%2B1%2Az=0%2C16%2Ax%2B4%2Ay%2B1%2Az=3%2C144%2Ax%2B12%2Ay%2B1%2Az=-5%29

  

  

  

  First let A=%28matrix%283%2C3%2C4%2C2%2C1%2C16%2C4%2C1%2C144%2C12%2C1%29%29. This is the matrix formed by the coefficients of the given system of equations.

  

  

  Take note that the right hand values of the system are 0, 3, and -5 and they are highlighted here: 

  

  

  

  

  These values are important as they will be used to replace the columns of the matrix A.

  

  

  

  

  Now let's calculate the the determinant of the matrix A to get abs%28A%29=-160. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.

  

  

  

  Notation note: abs%28A%29 denotes the determinant of the matrix A.

  

  

  

  ---------------------------------------------------------

  

  

  

  Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).

  

  

  

  

  

  

  Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=40. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.

  

  

  

  To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%2840%29%2F%28-160%29=-1%2F4

  

  

  

  So the first solution is x=-1%2F4

  

  

  

  

  ---------------------------------------------------------

  

  

  We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C4%2C2%2C1%2C16%2C4%2C1%2C144%2C12%2C1%29%29 again (this is the coefficient matrix).

  

  

  

  

  Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).

  

  

  

  

  

  

  Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=-480.

  

  

  

  To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%28-480%29%2F%28-160%29=3

  

  

  

  So the second solution is y=3

  

  

  

  

  ---------------------------------------------------------

  

  

  

  

  

  Let's reset again by letting A=%28matrix%283%2C3%2C4%2C2%2C1%2C16%2C4%2C1%2C144%2C12%2C1%29%29 which is the coefficient matrix.

  

  

  

  Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D 

  

  

  

  

  

  

  Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=800.

  

  

  

  To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28800%29%2F%28-160%29=-5

  

  

  

  So the third solution is z=-5

  

  

  

  

  ====================================================================================

  

  Final Answer:

  

  

  

  

  So the three solutions are x=-1%2F4, y=3, and z=-5 giving the ordered triple (-1/4, 3, -5)

  

  

  

  

  Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.