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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****Moderator**- Registered: 2005-06-28
- Posts: 20,892

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.

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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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