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anonymous
 3 years ago
The graph of f(x) = ax^4 + bx^2 + cx24 crosses the xaxis at 1, 2, and 3. determine the equation of f(x).
anonymous
 3 years ago
The graph of f(x) = ax^4 + bx^2 + cx24 crosses the xaxis at 1, 2, and 3. determine the equation of f(x).

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0start with \[(x1)(x+2)(x3)(xr)\] and see if you can solve for \(r\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do i expand and then solve?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no you don't really have to expand yet, you can visualize what the constant should be the constant will be \(6r\) so you know \(6r=24\) and if you are lucky \(r=4\) works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh damn it doesn't because you get a cube term and your polynomial doesn't seem to have one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0on the other hand something is wrong, because you have a 4th degree polynomial. it has 4 zeros total, although either two or four could be complex. however, if it crosses the \(x\) axis 3 times it must cross it a fourth. if one root is repeated, then it does not cross the axis but only touches it. so i am stumped

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe one passes through origin?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then it still has 4 zeros one would be zero, and also it could not end in \(24\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it really starts with \(ax^4\) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have the answer if that helps?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i am very certain that a fourth degree polynomial cannot cross the \(x\) axis only 3 times

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure go ahead and let me see why i am wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its f(x)= 4x^3 + 8x^2 + 20 x 20

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we could perhaps start with \[a(x1)(x+2)(x3)(xr)\] and see if we can solve for the coefficients, but it must cross 4 times

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(x)= 4x^3 + 8x^2 + 20 x 24 *** sorrry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold the phone, it is cubic?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answers a cubic :S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought it was a 4th degree polynomial, since you wrote \[f(x) = ax^4 + bx^2 + cx24\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if it is \[f(x) = ax^3 + bx^2 + cx24\] then that is a whole different story

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol thats what the question says, its ax^4 + bx^2 ..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and much simpler write \[a(x1)(x3)(x+2)= ax^3 + bx^2 + cx24\] and solve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then there is something wrong with the question, it is a typo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so if its ax^3 will the answer be right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your question asks for a 4th degree polynomial, which must have 4 zeros. it also has no cube term your answer has a third degree polynomial, so there is a mistake in the question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah it prob is a typo.. ah been trying to question for so long.. thanks tho!
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