please help!! Given three points, say (1, 1), (2, 3), and (-1, -2), in the plane

Question

please help!!

Given three points, say (1, 1), (2, 3), and (-1, -2), in the plane, find a parabola f(x) = ax^2 + bx + c whose graph contains all three of the points. a. Again, the idea is to solve a system of linear equations. In this case, there are exactly 1 such parabolas. One such parabola is f(x) = ax^2 + bx + c where a = b = and c = b. This is called an interpolation problem. Given k points (x_1, y_1), (x_2, y_2), ..., (x_k, y_k) where the x-coordinates are all different, you will see later in this course that there is a unique polynomial of degree at most k-1 whose graph contains all k of these points. This is a generalization of the well known fact that through any two distinct points you can always pass exactly one line. At this point in the course, if you are given a specific set of k points, you should know how to find such a polynomial.

Explanation / Answer

Ans A):

Given parabola f(x)= ax2+bx+c and points (1,1),(2,3) and (-1,-2)

Assuming the parabola is up-down, it has form of y=ax2+bx+c

now substitute points in equation.

For the first point (1,1) :

1=a(1)^2+b(1)+c

this reduces to

1=a+b+c

solve for c and we have

c=1-a-b --- > (equation i)

For second point (2,3)

3=a(2)^2+b(2)+c

this reduced to

3=4a+2b+c

solve for c and we have

c=3-4a-2b ----- (equation ii)

Before we do the third point, let's combine these equations, eliminating C:

3-4a-2b=1-a-b

-3a-b=-2

3a+b=2

b=2-3a ---- (call this as equation iii)

For point (-1,-2):

-2=a(-1)^2+b(-1)+c

this reduced to

-2=a-b+c

solve for b and we have

b=a+c+2

now substitute c value from equation (i)

b=a-a-b+1+2

2b=3

b=3/2

now substitute b value in equation (iii)

b=2-3a

3/2=2-3a

3a=2-3/2=1/2

a=1/6

now substitiute a, b values in equation (i):

c=-a-b+1

c=-1/6-3/2+1

c=-10/6+1

c=-5/3+1

c=-2/3

So the quadratic equation for parabola passing through given three point is y=1/6x^2+3/2x-2/3

So a = 1/6 , b = 3/2 , c=-2/3

Lets's test our answer by inserting given points to obtained quadratic equation.

For point(1,1):

1=1/6(1)^2+3/2(1)-2/3

1=1/6+3/2-2/3

1=5/3-2/3

1=3/3

1=1 ( Correct)

for point (2,3):

3=1/6(2)^2+3/2(2)-2/3

3=4/6+3-2/3

3=2/3+3-2/3

3=3 (correct)

For point (-1,-2):

-2=1/6(-1)^2+3/2(-1)-2/3

-2=1/6-3/2-2/3

-2=-8/6-2/3

-2=-4/3-2/3

-2=-6/3

-2=-2 ( Correct)

b) Solution :

I think you want to fit an order n polynomial to n points. Here's the method. I'll illustrate with a fourth-degree polynomial, although it can be used for any nth-degree polynomial.

Suppose you know the following points (x,f(x)):

(-2,85)
(-1,-8)
(1,-20)
(3,40)
(4,307)

There is some polynomial,

f(x) = ax4 + bx3 + cx2 + dx + e

that satisfies these five points. Since that's true, you can take any point -- I'll use the first one -- and set x equal to -2, and f(x) is 85. So you have

85 = a(-2)4 + b(-2)3 + c(-2)2 + d(-2) + e

Simplifying, we have

85 = 16a - 8b + 4c - 2d + e

Using the same procedure with the second point, we get

-8 = a - b + c - d + e

Doing the same calculations on the remaining three points, we get

-20 = a + b + c + d + e
40 = 81a + 27b + 9c + 3d + e
307 = 256a + 64b + 16c + 4d + e

Now here's something unexpected: we have five linear equations with five unknowns. The surprise is that the unknown isn't x, but rather a, b, c, d, and e. These five letters normally represent constant coefficients, but in this case, we don't know what they are. So to us, they're unknowns. x, on the other hand, is perfectly well known, at least for the five points we were given.

Make sure you have gotten past the shock of the knowns and unknowns switching places before you go on.

Are you OK? Let's continue.

So we have five equations with five unknowns:

16a - 8b + 4c - 2d + e = 85
a - b + c - d + e = -8
a + b + c + d + e = -20
81a + 27b + 9c + 3d + e = 40
256a + 64b + 16c + 4d + e = 307

Cramer's method works well for solving these (especially if you have a darned good method for finding determinants of large matrices!)

The coefficient matrix has a determinant of 21600.

When the first column of the coefficient matrix is replaced by the constants, the determinant becomes 64800, so

a=64800/21600=3.

By replacing each column of the coefficient matrix in turn by the constants, we find the values of b, c, d, and e the same way.

b=-7, c=0, d=1, and e=-17.

This is a good method for finding all sorts of curves given a number of points. For example, you can use a similar method to find the equation of a circle that passes through any three points. The only difference is the general form of the equation should be the one for a circle instead of the one for a polynomial.

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