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

• MHB
• takelgith
In summary: The answer is, "you don't". Because this is an even function, 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/
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

Where is the attempted work for this problem?

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 - - -

Cbarker1 said:
Where is the attempted work for this problem?

I just posted it as a reply to the forum

takelight said:
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 $b$.

takelight said:
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.

takelight said:
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

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

-Dan

## 1. How do you solve a quartic polynomial with a symmetric graph and intercept -2?

To solve a quartic polynomial with a symmetric graph and intercept -2, you can follow these steps:

1. Write the polynomial in standard form, with the highest degree term first and all terms in descending order.

2. Use the quadratic formula to find the roots of the polynomial.

3. Use the roots to factor the polynomial into two quadratic equations.

5. The solutions to the quartic polynomial will be the combination of the solutions to the two quadratic equations.

## 2. Can a quartic polynomial with a symmetric graph and intercept -2 have complex solutions?

Yes, a quartic polynomial with a symmetric graph and intercept -2 can have complex solutions. This means that the solutions will involve imaginary numbers.

## 3. What does it mean for a polynomial to have a symmetric graph?

A polynomial has a symmetric graph if it is symmetrical about the y-axis. This means that the left and right sides of the graph are mirror images of each other.

## 4. How do you determine the number of solutions for a quartic polynomial with a symmetric graph and intercept -2?

The number of solutions for a quartic polynomial with a symmetric graph and intercept -2 can be determined by looking at the degree of the polynomial. A quartic polynomial has a degree of 4, so it can have up to 4 solutions.

## 5. Can you use the graph of a quartic polynomial with a symmetric graph and intercept -2 to find its solutions?

Yes, you can use the graph of a quartic polynomial with a symmetric graph and intercept -2 to estimate the solutions. The x-intercepts of the graph will give you the solutions to the polynomial.

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