SOLUTION: Given 3 points f(3)= -11 f(1)= -3 f(-2)= -21 FIND THE QUADRATIC FUNCTION(AX^2*BX+C=0)

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Question 448999: Given 3 points
f(3)= -11
f(1)= -3
f(-2)= -21
FIND THE QUADRATIC FUNCTION(AX^2*BX+C=0)
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

find the Quadratic(ax^2 + bx + c = 0), Given 3 points

f(3)= -11

f(1)= -3

f(-2)= -21

 9a + 3b + c = -11 

  a + b + c = -3

 4a  -2b + c = -21

  a = -2, b = 4, c = -5

 -2x^2 +4x -5 = 0






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




  

  

  

  

  

  

  First let A=%28matrix%283%2C3%2C9%2C3%2C1%2C1%2C1%2C1%2C4%2C-2%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 -11, -3, and -21 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=30. 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=-60. 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=%28-60%29%2F%2830%29=-2

  

  

  

  So the first solution is x=-2

  

  

  

  

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  We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C9%2C3%2C1%2C1%2C1%2C1%2C4%2C-2%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=120.

  

  

  

  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=%28120%29%2F%2830%29=4

  

  

  

  So the second solution is y=4

  

  

  

  

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

  

  

  

  

  

  Let's reset again by letting A=%28matrix%283%2C3%2C9%2C3%2C1%2C1%2C1%2C1%2C4%2C-2%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=-150.

  

  

  

  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=%28-150%29%2F%2830%29=-5

  

  

  

  So the third solution is z=-5

  

  

  

  

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

  

  Final Answer:

  

  

  

  

  So the three solutions are x=-2, y=4, and z=-5 giving the ordered triple (-2, 4, -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.