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

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satellite73Best ResponseYou've already chosen the best response.1
start with \[(x1)(x+2)(x3)(xr)\] and see if you can solve for \(r\)
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
do i expand and then solve?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
no 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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oh damn it doesn't because you get a cube term and your polynomial doesn't seem to have one
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
on 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
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
maybe one passes through origin?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
then it still has 4 zeros one would be zero, and also it could not end in \(24\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
it really starts with \(ax^4\) right?
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
i have the answer if that helps?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
ok i am very certain that a fourth degree polynomial cannot cross the \(x\) axis only 3 times
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
sure go ahead and let me see why i am wrong
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
its f(x)= 4x^3 + 8x^2 + 20 x 20
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
we 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
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
f(x)= 4x^3 + 8x^2 + 20 x 24 *** sorrry
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
hold the phone, it is cubic?
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
the answers a cubic :S
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i thought it was a 4th degree polynomial, since you wrote \[f(x) = ax^4 + bx^2 + cx24\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
if it is \[f(x) = ax^3 + bx^2 + cx24\] then that is a whole different story
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
lol thats what the question says, its ax^4 + bx^2 ..
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and much simpler write \[a(x1)(x3)(x+2)= ax^3 + bx^2 + cx24\] and solve
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
then there is something wrong with the question, it is a typo
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
so if its ax^3 will the answer be right?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
your 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
 one year ago

sabika13Best ResponseYou've already chosen the best response.0
yeah it prob is a typo.. ah been trying to question for so long.. thanks tho!
 one year ago
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