Posted by Shawn on Wednesday, January 15, 2014 at 8:02am.
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.
Hmmm. In general, you cannot fit a quadratic to 4 points. Only a single cubic, but many of higher degree.
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
Oww. My Bad!