## Carissa15 one year ago Find the interpolating polynomial p(x)=a0+a1x+a2x^2 for the data (1,12),(2,15), (3,16). Can anyone help me with this question please?

1. anonymous

you will get 3 equations in three unknowns if you plug in

2. Carissa15

I do not understand how to even attempt this question :-(

3. anonymous

12=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2

4. anonymous

Do you see what i did?

5. Carissa15

oh ok, just by plugging in the values given for x and the result. Then what needs to be done with the equations?

6. anonymous

solve for a0, a1, a2, then plug them into the original one to get the result

7. anonymous

12=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2 simplified ; 12=a0+a1*1+a2*1 15=a0+a1*2+a2*4 16=a0+a1*3+a2*9 we have 3 equations in 3 unknowns, a0, a1, a2

8. Carissa15

solve for the value of a in each equation and then "check" that value in each equation?

9. anonymous

this is like solving 12=x+1y+1z 15=x+2y+4z 16=x+3y+4z

10. Carissa15

oh ok, so it will be very similar to my previous linear equations, solve for each unknown and sub back into original equation to see whether the equation has 1, many or no solutions?

11. ganeshie8

Hey!

12. Carissa15

hey

13. ganeshie8

There is an easier way to find the interpolation polynomial, wana learn ?

14. Carissa15

sure, that would be great

15. ganeshie8

so you want to find a quadratic that passes through the points (1,12),(2,15), (3,16)

16. ganeshie8

Look at below quadratic, notice anything interesting ? $\dfrac{(x-2)(x-3)}{(1-2)(1-3)}\cdot 12$

17. ganeshie8

what is its value when $$x=1$$ ?

18. Carissa15

12?

19. ganeshie8

Yes, what do you get when you plugin $$x=2$$ or $$x = 3$$ ?

20. misty1212

this looks interesting, don't know it @ganeshie8 got a source for this i can read?

21. ganeshie8

one sec, i have a really nice short pdf on this...

22. misty1212

cool thnx

23. Carissa15

I got 0

24. ganeshie8

here it is http://www2.lawrence.edu/fast/GREGGJ/Math420/Section_3_1.pdf @misty1212

25. ganeshie8

@Carissa15 rigjt, so we have just cooked up a quadratic that passes through $$(1, 12)$$ but produces $$0$$ at two other points that we are interested in

26. misty1212

great thanks!

27. ganeshie8

Here is the final quadratic that passes through $$(1,12),(2,15), (3,16)$$ : $\dfrac{(x-2)(x-3)}{(1-2)(1-3)}\cdot 12 + \dfrac{(x-1)(x-3)}{(2-1)(2-3)}\cdot 15 + \dfrac{(x-1)(x-2)}{(3-1)(3-2)}\cdot 16$ see if you can figure out why/how it works.. :)

28. Carissa15

each og the equations equal 1 and therefore are only multiplied by the 12,15 and 16. giving total 43?

29. ganeshie8

you don't have teamviewer?

30. Carissa15

no, i don't :(