Determining the zeros and end behavior of a function

[rachelmaddie]

Hi. I need my work checked please.

The zeros of a function are all the values of x for which f(x) = 0
Therefore, to find the zeros of the function:
equal f(x) to zero and solve for x.
f(x) = x(x - 4)(x + 2) = 0
x(x - 4)(x + 2) = 0

We have the multiplicity of 3 factors x,
(x - 4) and (x + 2)

The function will be equal to zero when one of the factors is equal to zero, that is:
x = 0
(x - 4) = 0, x = 4
(x + 2) = 0, x = -2

Note that f(x) = x(x - 4)(x + 2) is a cubic function of positive principle coefficient, the graph starts from -∞ and cuts to the x-axis at x = -2
Then decreases and cuts by the x-axis at x = 0
For the third time, it cuts the x-axis at x = 4 and then tends toward ∞
*The graph rises to the left and rises to the right.*

x = 0 (Multiplicity of 1)
x = 4 (Multiplicity of 1)
x = -2 (Multiplicity of 4)

 

[JeffM]

If I were grading this, I'd give you partial credit. There are several material errors despite some solid work.

The zeros of a function are all the values of x for which f(x) = 0

Correct.

Therefore, to find the zeros of the function:
equal f(x) to zero and solve for x.
f(x) = x(x - 4)(x + 2) = 0
x(x - 4)(x + 2) = 0

But actually the function is [MATH]x(x - 4)(x + 2)^4.[/MATH] So this is a technical error.

We have the multiplicity of 3 factors x,
(x - 4) and (x + 2)

This is an error in vocabulary. What we have are three distinct factors, one with a multiplicity of four.

The function will be equal to zero when one of the factors is equal to zero, that is:
x = 0
(x - 4) = 0, x = 4
(x + 2) = 0, x = -2

This is a little sloppy because (x + 2)^4 is the factor but it will be zero if and only if (x + 2) = 0 so no big deal.

Note that f(x) = x(x - 4)(x + 2) is a cubic function of positive principle coefficient, the graph starts from -∞ and cuts to the x-axis at x = -2
Then decreases and cuts by the x-axis at x = 0
For the third time, it cuts the x-axis at x = 4 and then tends toward ∞
*The graph rises to the left and rises to the right.*

Now your mistake about the exponent comes to bite you. f(x) is not a polynomial of degree 3; it is a polynomial of degree 6, which is far more complicated than a polynomial of degree 3. I doubt you have the tools to determine exactly where the sign of the slope changes, but because it is a polynomial of even degree with a positive leading coefficient, you can determine that the graph starts and ends at positive infinity. This is a big enough error that some teachers might give you no credit.

x = 0 (Multiplicity of 1)
x = 4 (Multiplicity of 1)
x = -2 (Multiplicity of 4)

Correct.

 

[rachelmaddie]

If I were grading this, I'd give you partial credit. There are several material errors despite some solid work.

Correct.

But actually the function is [MATH]x(x - 4)(x + 2)^4.[/MATH] So this is a technical error.


This is an error in vocabulary. What we have are three distinct factors, one with a multiplicity of four.


This is a little sloppy because (x + 2)^4 is the factor but it will be zero if and only if (x + 2) = 0 so no big deal.


Now your mistake about the exponent comes to bite you. f(x) is not a polynomial of degree 3; it is a polynomial of degree 6, which is far more complicated than a polynomial of degree 3. I doubt you have the tools to determine exactly where the sign of the slope changes, but because it is a polynomial of even degree with a positive leading coefficient, you can determine that the graph starts and ends at positive infinity. This is a big enough error that some teachers might give you no credit.


Correct.

My question is how could I better word the last part for the graph?

 

[JeffM]

If you do not know differential calculus, I would not even try to identify turning points of a polynomial of degree six: there is at least one and may be as many as five. Even if you do know differential calculus, finding the turning points requires finding the roots of a quintic, which cannot generally be done using algebra. But the question does not ask you to find the turning points. You imposed that burden on yourself. It merely asks you to describe the end behavior.

My opinion, which may not be the same as your teacher, is that it is a good answer to say

"As a polynomial of even degree with a positive leading coefficient, the end behavior of the function for negative x and for positive x is to approach positive infinity."

I'd be happy if others would chime in with their thoughts.

 

[rachelmaddie]

If you do not know differential calculus, I would not even try to identify turning points of a polynomial of degree six: there is at least one and may be as many as five. Even if you do know differential calculus, finding the turning points requires finding the roots of a quintic, which cannot generally be done using algebra. But the question does not ask you to find the turning points. You imposed that burden on yourself. It merely asks you to describe the end behavior.

My opinion, which may not be the same as your teacher, is that it is a good answer to say

"As a polynomial of even degree with a positive leading coefficient, the end behavior of the function for negative x and for positive x is to approach positive infinity."

I'd be happy if others would chime in with their thoughts.

How does this look?
The zeros of a function are all the values of x for which f(x) = 0
Therefore, to find the zeros of the function:
equal f(x) to zero and solve for x.
f(x) = x(x - 4)(x + 2)^4 = 0
x(x - 4)(x + 2)^4 = 0

We have three distinct factors, one with a multiplicity of four.
(x - 4) and (x + 2)

(x + 2)^4 is the factor but the function will be equal to zero if and only if (x + 2) = 0, that is:
x = 0
(x - 4) = 0, x = 4
(x + 2) = 0, x = -2

As a polynomial of even degree with a positive leading coefficient, the end behavior of the function for negative x and for positive x is to approach positive infinity.

x = 0 (Multiplicity of 1)
x = 4 (Multiplicity of 1)
x = -2 (Multiplicity of 4)

 

[JeffM]

How does this look?
The zeros of a function are all the values of x for which f(x) = 0
Therefore, to find the zeros of the function:
equal f(x) to zero and solve for x.
f(x) = x(x - 4)(x + 2)^4 = 0
x(x - 4)(x + 2)^4 = 0

We have three distinct factors, one with a multiplicity of four.
(x - 4) and (x + 2)

(x + 2)^4 is the factor but the function will be equal to zero if and only if (x + 2) = 0, that is:
x = 0
(x - 4) = 0, x = 4
(x + 2) = 0, x = -2

As a polynomial of even degree with a positive leading coefficient, the end behavior of the function for negative x and for positive x is to approach positive infinity.

x = 0 (Multiplicity of 1)
x = 4 (Multiplicity of 1)
x = -2 (Multiplicity of 4)

It looks fine to me, but the important question is whether you fully understand why the revisions were necessary.