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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?
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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you will get 3 equations in three unknowns if you plug in

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0I do not understand how to even attempt this question :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.012=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you see what i did?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0oh ok, just by plugging in the values given for x and the result. Then what needs to be done with the equations?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve for a0, a1, a2, then plug them into the original one to get the result

anonymous
 one year ago
Best ResponseYou've already chosen the best response.012=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

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0solve for the value of a in each equation and then "check" that value in each equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is like solving 12=x+1y+1z 15=x+2y+4z 16=x+3y+4z

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0oh 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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0There is an easier way to find the interpolation polynomial, wana learn ?

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0sure, that would be great

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so you want to find a quadratic that passes through the points (1,12),(2,15), (3,16)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Look at below quadratic, notice anything interesting ? \[\dfrac{(x2)(x3)}{(12)(13)}\cdot 12\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what is its value when \(x=1\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes, what do you get when you plugin \(x=2\) or \(x = 3\) ?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0this looks interesting, don't know it @ganeshie8 got a source for this i can read?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0one sec, i have a really nice short pdf on this...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0here it is http://www2.lawrence.edu/fast/GREGGJ/Math420/Section_3_1.pdf @misty1212

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0@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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Here is the final quadratic that passes through \((1,12),(2,15), (3,16)\) : \[\dfrac{(x2)(x3)}{(12)(13)}\cdot 12 + \dfrac{(x1)(x3)}{(21)(23)}\cdot 15 + \dfrac{(x1)(x2)}{(31)(32)}\cdot 16\] see if you can figure out why/how it works.. :)

Carissa15
 one year ago
Best ResponseYou've already chosen the best response.0each og the equations equal 1 and therefore are only multiplied by the 12,15 and 16. giving total 43?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you don't have teamviewer?
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