MHB Solving a Quartic Polynomial with Symmetric Graph & Intercept -2

[takelgith]

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

 

[cbarker1]

Where is the attempted work for this problem?

 

[takelgith]

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 - - -

Where is the attempted work for this problem?

I just posted it as a reply to the forum

 

[Opalg]

f(x)=ax^4+bx^2−2

.
.
.

f′(x)=4ax^3+2bx=0 at x=±2

From here how do I find a and b to finish the function...?

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 $b$.

 

[HallsofIvy]

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

Yes, that's good. You want to find values of the three parameters, a, b, and c so you need three equations.

y-intercept = -2
⟹c=−2

Yes, a(0^4)+ b(0^2)+ c= c= -2 is one of the equations.

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

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.

 

[skeeter]

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

Help solving this?? - My Math Forum

 

[topsquark]

Help solving this?? - My Math Forum

Don't'cha just love cross posting? If only they knew that a significant number of members are on many of these boards!

-Dan