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?

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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?

Mathematics
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you will get 3 equations in three unknowns if you plug in
I do not understand how to even attempt this question :-(
12=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2

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Do you see what i did?
oh ok, just by plugging in the values given for x and the result. Then what needs to be done with the equations?
solve for a0, a1, a2, then plug them into the original one to get the result
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
solve for the value of a in each equation and then "check" that value in each equation?
this is like solving 12=x+1y+1z 15=x+2y+4z 16=x+3y+4z
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?
Hey!
hey
There is an easier way to find the interpolation polynomial, wana learn ?
sure, that would be great
so you want to find a quadratic that passes through the points (1,12),(2,15), (3,16)
Look at below quadratic, notice anything interesting ? \[\dfrac{(x-2)(x-3)}{(1-2)(1-3)}\cdot 12\]
what is its value when \(x=1\) ?
12?
Yes, what do you get when you plugin \(x=2\) or \(x = 3\) ?
this looks interesting, don't know it @ganeshie8 got a source for this i can read?
one sec, i have a really nice short pdf on this...
cool thnx
I got 0
here it is http://www2.lawrence.edu/fast/GREGGJ/Math420/Section_3_1.pdf @misty1212
@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
great thanks!
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.. :)
each og the equations equal 1 and therefore are only multiplied by the 12,15 and 16. giving total 43?
you don't have teamviewer?
no, i don't :(

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