SOLUTION: Find the quadratic polynomial whose graph passes through the points (1,14) (2,28) (3,48)

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Question 1161075: Find the quadratic polynomial whose graph passes through the points (1,14) (2,28) (3,48)
Found 3 solutions by MathLover1, solver91311, greenestamps:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
use given point (,)

......solve for
.........eq.1
use given point (,)

......solve for
.........eq.2
from eq.1 and eq.2 we have
.....solve for

.........eq.3
go to
.........eq.1, substitute


...............eq.4

go to
use given point (,), substitute from eq.3 and from eq.4
........solve for
.....simplify




go to
.........eq.3, substitute


go to
...............eq.4, substitute


your equation is:








Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


A quadratic polynomial function has the form:



Any point on the graph of this function has the form



So, if the point is on the graph, then the following must hold:



Similarly, for and



and



must hold.

These three equations reduce to the 3X3 linear system: 
              a  +  b  +  c  =  14

             4a  + 2b  +  c  =  28

             9a  + 3b  +  c  =  48


Solve the system for the required , , and coefficients.


John

My calculator said it, I believe it, that settles it
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The standard method for solving a problem like this is as shown by the other two tutors: use the x and y coordinates of the given points in the standard equation to get three equations in the coefficients a, b, and c; then solve the system of equations.

Here is a completely different method that I find a bit less work. This alternative method can be used because the y values are given for three consecutive x values.

So think of the problem as a sequence of numbers produced by a quadratic polynomial.

In a sequence produced by the quadratic polynomial , the second differences are the constant 2a. (If you know some basic calculus, you can see that is true, since the second derivative of the polynomial is 2a.)

In this sequence, the second difference is 6:
   14  18  48
     14  20
        6

Since we now that second difference is 2a, we know that the leading coefficient a is 3, and the quadratic polynomial is



The "bx+c" is a linear polynomial; it can be determined easily by comparing the actual y values to the value of 3x^2:
   x   3x^2   y   y-3x^2 (= bx+c)
  -------------------------------
   1     3   14     11
   2    12   28     16
   3    27   48     21

It is easy to see that the values in the last column are bx+c = 5x+6.

So, already knowing that the leading term of the quadratic is 3x^2, we now know that the complete quadratic is
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