Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Do you see what i did?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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?

Answer 11

Hey!

Answer 12

hey

Answer 13

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

Answer 14

sure, that would be great

Answer 15

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

Answer 16

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

Answer 17

what is its value when \(x=1\) ?

Answer 18

12?

Answer 19

Yes, what do you get when you plugin \(x=2\) or \(x = 3\) ?

Answer 20

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

Answer 21

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

Answer 22

cool thnx

Answer 23

I got 0

Answer 24

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

@misty1212

Answer 25

@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

Answer 26

great thanks!

Answer 27

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.. :)

Answer 28

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

Answer 29

you don't have teamviewer?

Answer 30

no, i don't :(