Re: Finding equation of a parabola with the use of 3 points -- Mathematics Help

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Subject: Re: Finding equation of a parabola with the use of 3 points

Author:
QUITTNER

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Date Posted: 06:21:51 05/23/03 Fri
In reply to: Joanna 's message, "Finding equation of a parabola with the use of 3 points" on 23:51:55 05/11/03 Sun

Joanna, I assume that you know how to invert a matrix. One of he equations of a parabola is A+Bx+Cx^2=y, which can also be written as Ax^0+Bx^1+Cx^2=y, where x^0 is always 1. A Polynomial of nth degree needs n+1 points; here we have a 2nd degree polynomial and 3 points, that's OK.
..... Let matrix M1 equal the left side of the equation, and fill in all the x's as follows:
Take one of the points (in any order), (3,-1) first. Its x is 3, therefore the top line of M1 is 1, for x'0, 3 for x, and 9 for x'2. The 2nd line is for point (1,-7): 1,1,1. The 3rd line for point (-2,14) is 1,-2,4. Therefore, to repeat, M1 is

| 1 3 9 | for point 1,
| 1 1 1 } 2,
| 1 =2 4 | 3.

The right side of the equation of the parabola is only y, we then make Matrix M2 equal to
| -1 | for point 1,
| -7 | 2,
| 14 | 3.

Next invert M1 to get matrix M3:
| 1/5 1 1/5 |
| 1/10 1/6 -15/4 |
| 1/10 -1/6 1/15 |

To get the results matrix, multiply M2 by M3, giving
| -4 |
| -5 |
| 2 | Therefore the equation of the parabola is
A+Bx+Cx^2=y where A=-4, B=-5 and C=2. Make sure you check this by evaluating the equation for each of the given x values of each of the 3 points and see that you get the given y's.

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Replies:

[> [> Subject: Finding equation of a parabola with vertex at Oy and passing thru (2,2)

Author:
lanie
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Date Posted: 18:00:29 11/06/03 Thu

>Joanna, I assume that you know how to invert a matrix.
>One of he equations of a parabola is A+Bx+Cx^2=y,
>which can also be written as Ax^0+Bx^1+Cx^2=y, where
>x^0 is always 1. A Polynomial of nth degree needs n+1
>points; here we have a 2nd degree polynomial and 3
>points, that's OK.
>..... Let matrix M1 equal the left side of the
>equation, and fill in all the x's as follows:
>Take one of the points (in any order), (3,-1) first.
>Its x is 3, therefore the top line of M1 is 1, for
>x'0, 3 for x, and 9 for x'2. The 2nd line is for point
>(1,-7): 1,1,1. The 3rd line for point (-2,14) is
>1,-2,4. Therefore, to repeat, M1 is
>
>| 1 3 9 | for point 1,
>| 1 1 1 } 2,
>| 1 =2 4 | 3.
>
>The right side of the equation of the parabola is only
>y, we then make Matrix M2 equal to
>| -1 | for point 1,
>| -7 | 2,
>| 14 | 3.
>
>Next invert M1 to get matrix M3:
>| 1/5 1 1/5 |
>| 1/10 1/6 -15/4 |
>| 1/10 -1/6 1/15 |
>
>To get the results matrix, multiply M2 by M3, giving
>| -4 |
>| -5 |
>| 2 | Therefore the equation of the parabola is
>A+Bx+Cx^2=y where A=-4, B=-5 and C=2. Make sure
>you check this by evaluating the equation for each of
>the given x values of each of the 3 points and see
>that you get the given y's.

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[> [> [> Subject: Re: Finding equation of a parabola with vertex at Oy and passing thru (2,2)

Author:
QUITTNER
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Date Posted: 11:19:31 11/11/03 Tue

You wrote: "Finding equation of a parabola with vertex at Oy and passing thru (2,2)"
..... I suppose this means that the vertex is at the ORIGIN (0,0), and that one point is at (2,2)
..... We have the general equation y-k=a(x-h)^2 where, in our case k and h are both zero. Therefore
y=a*x^2 and in our case 2=a*2^2=4a. Therefore a=1/2 and
y=(1/2)*x^2
..... Check it out: if x=2 then y=1/2 of 4, y=2 OK

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