- #1
takelgith
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Find the equation of a quartic polynomial whose graph is symmetric about the y -axis and has local maxima at (−2,0) and (2,0) and a y -intercept of -2
Solving a Quartic Polynomial with Symmetric Graph & Intercept -2
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Discussion Overview
The discussion revolves around finding the equation of a quartic polynomial that is symmetric about the y-axis, has local maxima at the points (−2,0) and (2,0), and a y-intercept of -2. The focus is on the mathematical reasoning and calculations involved in determining the coefficients of the polynomial.
Discussion Character
- Mathematical reasoning
- Homework-related
- Technical explanation
Main Points Raised
- Some participants note that the polynomial must be an even function, expressed as f(x)=ax^4+bx^2+c, with the y-intercept condition leading to c=−2.
- Participants discuss the implications of local extrema at x=±2, leading to the equations f(2) = 0 and f'(2) = 0, which provide relationships between the coefficients a and b.
- One participant points out that the equations derived from the conditions at the local extrema yield two equations for a and b, specifically 16a + 4b - 2 = 0 and 32a + 4b = 0.
- Another participant suggests that due to the even nature of the function, the relationship between a and b can be expressed as b= -(16/3)a, indicating a family of functions that satisfy the given conditions.
- There is a repeated request for help in solving for a and b, indicating uncertainty in the process.
Areas of Agreement / Disagreement
Participants generally agree on the structure of the polynomial and the implications of the symmetry and local extrema. However, there is no consensus on the specific values of a and b, as the discussion reveals multiple approaches and some uncertainty regarding the completion of the solution.
Contextual Notes
Participants express uncertainty about the values of a and b, and the discussion highlights the dependence on the derived equations from the local extrema conditions. The exploration of the polynomial's properties remains unresolved.
- #2
cbarker1
Gold Member
MHB
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Where is the attempted work for this problem?
- #3
takelgith
- 6
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This is the attempted work for this. I don't know how to find a and b from this...
symmetric over the y-axis
⟹
⟹
f(x) is an even quartic function ...
f(x)=ax^4+bx^2+c
y-intercept = -2
⟹c=−2
f(x)=ax^4+bx^2−2
the local extrema info tells me two things...
f(x)=0
at
x=±2
f′(x)=4ax^3+2bx=0 at x=±2
From here how do I find a and b to finish the function...?
- - - Updated - - -
I just posted it as a reply to the forum
symmetric over the y-axis
⟹
⟹
f(x) is an even quartic function ...
f(x)=ax^4+bx^2+c
y-intercept = -2
⟹c=−2
f(x)=ax^4+bx^2−2
the local extrema info tells me two things...
f(x)=0
at
x=±2
f′(x)=4ax^3+2bx=0 at x=±2
From here how do I find a and b to finish the function...?
- - - Updated - - -
Cbarker1 said:
I just posted it as a reply to the forum
- #4
Opalg
Gold Member
MHB
- 2,778
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You know that $f(x)$ and $f'(x)$ are both $0$ at the points $(\pm2,0)$. This tells you that $$f(2) = 0:\qquad 16a + 4b - 2 = 0,$$ $$f'(2) = 0:\qquad 32a + 4b = 0.$$ There are your two equations for $a$ and $b$.takelight said:
- #5
HallsofIvy
Science Advisor
Homework Helper
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Yes, that's good. You want to find values of the three parameters, a, b, and c so you need three equations.takelight said:
Yes, a(0^4)+ b(0^2)+ c= c= -2 is one of the equations.
The answer is, "you don't". Because this is an even functions, any local extremum at x= a necessarily corresponds to a
local extremum
at x= -a. Putting either x= 2 or x= -2 into the equation gives the same thing:
f'(2)= 4a(2^3)+ 3b(2)= 32a+ 6b= 0.
f'(-2)= 4a((-2)^3+ 3b(-3)= -32a- 6b= 0.
From 32a+ 6b= 0, we get b= -(32/6)a= -(16/3)a so the best we can say is that any function of the form f(x)= ax^4- (16/3)ax^2- 2 satisfies all of those conditions.
- #6
skeeter
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- #7
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Don't'cha just love cross posting? If only they knew that a significant number of members are on many of these boards!skeeter said:
-Dan
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