EMERGENCY! I REALLY NEED HELP ON THIS QUESTION! WILL MEDAL!
HERE IS THE QUESTION
@vera_ewing @Howard-Wolowitz @paki @ganeshie8 @just_one_last_goodbye @Calcmathlete @phi @tkhunny
@amistre64 please help!
whats the trouble?
I don't understand how to do it
I know I have to place the points in, but I dont know which parts go where to solve it
what part of it DO you understand ... what is a multiplicity?
a multiplicity is what you multiply by?
it goes in the front like y=muultiplicity _____
hmm, no, it simply means there is more than one of the parts that is affected
take (x+b)^2 and write it as: (x+b)(x+b) y = a (x+b) (x+b) (x-c) now when x=-1, only 2 of these parts zero out .... which 2 do you think they would be?
if x=-1, wouldn't the x terms cancel? and get turned to -1
if x=-1 then yes, the xs become -1 0 = a (-1+b) (-1+b) (-1-c) we are told that only 2 parts of this zero out ... a multiplicity of 2 ... when x=-1 which parts would you say zero out?
Okay, then the b terms would zero out?
yes, there are 2 identical parts ... x+b .. it stands to reason that these zero out -1+b = 0, what is the value of b?
b would be 1
good we have one variable solved for y = a(x+1)^2 (x-c) now, we can use the other point values, what are they?
(4,0) and (0, -12)
so, y=0 when x=4 is also a root lets use it 0 = a(4+1)^2 (4-c) what can we determine from this? what is the value of c?
ummm c would be 4 and a would be 0
a is not 0, a is unknown at the moment ... c=4 zeros out so lets use that y = a(x+1)^2 (x-4) now we can use the last point, when x=0, y=-12 -12 = a(0+1)^2 (0-4) now solve for a
if a was 0, then y would always be zero regardless of the value for x, since 0 times anything else is zero
okay a would be 3
a=3 is correct
so the asnwer is 3?
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