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Mathematics 19 Online
OpenStudy (anonymous):

Write the function f(x)=x^3-2x^2+3x+5 as a polynomial of x-2

OpenStudy (jamesj):

Well, it must be third order. So expand it out term by term. Start with (x-2)^3 and then see what you have left i.e., write f(x) = (x-2)^3 + stuff. Once you've figured out what stuff is, you'll be able to find the coefficient A such that f(x) = (x-2)^3 + A(x-2)^2 + other stuff now repeat that step and find the coefficient B such that f(x) = (x-2)^3 + A(x-2)^2 + B(x-2) + constant

OpenStudy (anonymous):

thank again @JamesJ

OpenStudy (anonymous):

jamesj is of course correct, but if this is for a calc class you can do this: \[f(x)=a(x-2)^3+b(x-2)^2+c(x-2)+d\] \[f(2)=d\] so you can replace \(x\) by 2 in the original expression and find \(d\) immediately then \[f'(x)=3a(x-2)^2+2b(x-2)+c\] and therefore \[f'(2)=c\] replace \(x\) by to in derivative of original expression to get \(c\) lather, rinse, repeat haven't seen @jamesj in a while. where you been?

OpenStudy (jamesj):

sat73, how are you? I've been busy! Nice method by the way.

OpenStudy (anonymous):

good, and glad you are busy, although we (i at least) missed you here

OpenStudy (jamesj):

thanks. I'll drop in now and again, but I have a few things cooking.

OpenStudy (anonymous):

great see you around the pool as they say

OpenStudy (jamesj):

that sounds good. I'll have Mai Tai please!

OpenStudy (anonymous):

-.-

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