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#1 2006-02-26 22:41:39

Titus
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Quadratics Again..

If a quadratic line passes through (1,-2) and (2,-14)

what is the line that satisfies this information?

are there any others that satisfy this? how many?

thanks smile

#2 2006-02-26 22:56:12

ganesh
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Re: Quadratics Again..

The line passing through (1, -2) and (2, -14) is given by
(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1). Here,
(y + 2)/-12 = (x - 1)/1
y+2 = -12x+12
y+12x = 10
or 12x + y = 10.
This is not a quadratic, it is a straight line.
This is the only equation of the line that passes through the two points.


Character is who you are when no one is looking.

#3 2006-02-27 01:11:09

Ricky
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Re: Quadratics Again..

We have y = ax^2 + bx + c.

Plugging in (1, -2), we get:

-2 = a + b + c

And plugging in (2, -14) we get:

-14 = 4a + 2b + c

Let's try as best we can to "solve" one of these.  The first one, when multipled by -2, becomes:

4 = -2a - 2b - 2c

Adding this to the second one, we get:

-10 = 2a - c

And this seems to be the only restriction.  So let's choose an a and c that works in the above equation, and then solve for b.

-10 = 2(-7) - (-4) = -10

So a = -7 and c = -4.  Putting these back into the very first equation:

-2 = -7 + b - 4
b = 9

a = -7 b = 9 c = -4.  But do these work?

-2 = -7(1)^2 + 9(1) - 4 which is true, so it goes through (1,-2)
-14 = -7(2)^2 + 9(2) - 4 which is true, so it goes through (2, -14)

But keep in mind, there are infinitely many solutions.  This is just one of them.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

#4 2006-02-27 12:22:25

naturewild
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Re: Quadratics Again..

That's true. You can draw your quadraic anyhow and you can still make it pass through those 2 points. A straight line, a third degree function, a fourth degree function etc..

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