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**Titus****Member**- Registered: 2005-03-07
- Posts: 10

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

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,368

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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

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**naturewild****Member**- Registered: 2005-12-04
- Posts: 30

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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