Posted by **Shawn** on Wednesday, January 15, 2014 at 8:02am.

I am trying to complete this input/output table. Input 0 output blank, input 3/2 output 17/6, input 11/4 output blank, input 3 output 13/3, input blank output 13/2

- Math -
**Reiny**, Wednesday, January 15, 2014 at 8:31am
Your relation could be linear:

using (3/2 , 17,6) and (3, 13/3)

slope = (17/6 - 13/3)/(3/2 - 3) = (-3/2) / (-3/2) = 1

y - 13/3 = 1(x-3)

3y - 13 = 3x-9

3x - 3y = -4

now plug in the given input as x or the given output as y

e.g input = x = 11/4

3(11/4) - 3y = -4

times 4

33 - 12y = -16

-12y = -49

y = -49/-12 = 49/12

do the remaining one in the same way

your relation could have been quadratic:

let's make (3,13/3) the vertex

then

y = a(x-3)^2 + 13/3

but (3/2 , 17/6) lies on it

17/6 = a(3/2 - 3)^2 + 13/3

17/6 = a(9/4) + 13/3

times 12

34 = 27a + 52

27a = -18

a = -18/27 = 2/3

so it could have been

y = (2/3)(x-3)^2 + 13/3

There could have been an infinite number of possible quadratics, not to mention, we could have made it into a cubic, etc.

- Math -
**Steve**, Wednesday, January 15, 2014 at 11:24am
Hmmm. In general, you cannot fit a quadratic to 4 points. Only a single cubic, but many of higher degree.

- Math -
**Reiny**, Wednesday, January 15, 2014 at 11:32am
Steve, only 2 points were actually given.

(3/2, 17/6)

(3, 13/3)

The others were

(0, ??)

(11/4, ??)

So I could form a unique linear, but an infinite number of quadratics, cubics, etc that pass through those two points.

Once I have an equation, the missing y values of the other 2 points are then found

- Reiny - Oww - my Bad! -
**Steve**, Wednesday, January 15, 2014 at 12:35pm
Oww. My Bad!

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