Relationship between x and y: (1,-1) (2,1) (3,3) I got y=2x-3

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Whats the relationship between x and y

(1,-1) (2,1) (3,3) I got y=2x-3

(1,2) (2,5) (3,10) I dont know how to do


many thanks
 
Whats the relationship between x and y

(1,-1) (2,1) (3,3)
Unless the instructions specified the type of relationship (for instance, "a linear function"), there may be many valid answers.

I got y=2x-3
You can confirm validity by plugging the x-values into your formula, and confirming that it returns the expected y-values.

(1,2) (2,5) (3,10) I dont know how to do
Did you plot the points? What sort of curve looks like it might fit through these points? A straight or a curved line? If a curved line, what sort of equation (that is, what degree of function) can you fully determine with three points?

If you get stuck, please reply showing your thoughts and efforts so far. Thank you! ;)
 
Whats the relationship between x and y

(1,-1) (2,1) (3,3) I got y=2x-3

(1,2) (2,5) (3,10) I dont know how to do


many thanks
In the first example you were lucky in that differences were the same, that is
\(\displaystyle \frac{1-(-1)}{2-1}\, =\, \frac{3-1}{3-2}\, =\, 2\).
Since that is the case, you get a linear function as one relationship between x and y [there are others which will work].

In the second case, you are not so lucky. Fortunately there is a 'formula' you can use called the Lagrange Interpolating Polynominal, see
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
for example. Take your case (1) for example. You would get
y = \(\displaystyle \frac{(x\, -\, 3)\, (x\, -\, 2)\, (-1)}{(1\, -\, 3)\, (1\, -\, 2)}\, +\, \frac{(x\, -\, 3)\, (x\, -\, 1)\, (1)}{(2\, -\, 3)\, (2\, -\, 1)}\, +\, \frac{(x\, -\, 1)\, (x\, -\, 2)\, (3)}{(3\, -\, 1)\, (3\, -\, 2)}\)
= \(\displaystyle \frac{-\, (x\, -\, 3)\, (x\, -\, 2)}{2}\, -\, \frac{2\, (x\, -\, 3)\, (x\, -\, 1)}{2}\, +\, \frac{3\, (x\, -\, 1)\, (x\, -\, 2)\, (3)}{2}\)
= \(\displaystyle \frac{1}{2}\, [-\, (x\, -\, 3)\, (x\, -\, 2)\, -\, 2\, (x\, -\, 3)\, (x\, -\, 1)\, +\, 3\, (x\, -\, 1)\, (x\, -\, 2)\, (3)]\)
= \(\displaystyle \frac{1}{2}\, [-\, x^2\, +\, 5x\, -\, 6\,\,\, -\, 2x^2\, +\, 8x\, -\, 6\,\,\, +\, 3x^2\, -\, 9x\, +\, 6]\)
= 2x - 3

Now do the same type of thing for (2)

EDIT: You can do the same sort of thing a different way. Again, we'll do your example (1): Start with the first point (1,-1)
y(x) = -1 + (x-1) g1(x)
Add the second point (2,1)
y(2) = 1 = -1 + (2-1) g1(2) = -1 + g1(2)
So g1(2) = 2 and
y(x) = -1 + (x-1) [ 2 + (x-2) g2(x) ]
Add the third point (3,3)
y(3) = 3 = -1 + (3-1) [ 2 + (3-2) g2(3) ] = -1 + 2 [ 2 + g2(3) ] = 3 + 2 g2(3)
So g2(3) = 0 and
y(x) = -1 + (x-1) [ 2 + (x-2) [0 + (x-3) g3(x)] ]
Which we can write as
y(x) = 2x - 3 + (x-1) (x-2) (x-3) g3(x)
Since we have all three points, we can take g3(x) as what ever we want. The 'obvious' value is g3(x)=0.
 
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