Polynomial graph, stuck

[christieeebee]

I have 4 points for a graph and i've spend 3 hours trying to figure it out.
my answer is p(x)=-1/3(x^3+x^2-9x-9)
the issue is both sides are supposed to be going down and unfortunately they're not.

 

[Deleted member 4993]

I have 4 points for a graph and i've spend 3 hours trying to figure it out.
my answer is p(x)=-1/3(x^3+x^2-9x-9)
the issue is both sides are supposed to be going down and unfortunately they're not.View attachment 17211

Please share your work/thoughts about this assignment. Things to consider:

The curve should be a fourth order polynomial (to replicate the desired shape) and

The x-axis is a tangent to the curve at x = -3 (local maximum y = 0)

 

[Dr.Peterson]

I have 4 points for a graph and i've spend 3 hours trying to figure it out.
my answer is p(x)=-1/3(x^3+x^2-9x-9)
the issue is both sides are supposed to be going down and unfortunately they're not.

In order to be falling in both directions, the degree should be even; and for the shape you show, it should be at least 4. (A polynomial with the actual shape you show would have a higher degree, because of the odd wiggles, but presumably it is only the given points and the overall behavior that is to be replicated!)

Focus on the factors, not the terms. That is, first write it in factored form based on the zeros and their multiplicities, and then determine the coefficient of the leading term to obtain the needed y-intercept.

Presumably you have been taught this, and should have seen examples before being given this exercise.

 

[JeffM]

First, there are an infinite number of polynomials with graphs that run through four given points. There is, however, only one polynomial of degree three with a graph that runs through four given points.

Second, only polynomials of even degree have slopes with the same sign at large absolute values of x. So you can't fit a polynomial of degree 3 to the given graph.

Third, the maximum number of local extrema possible for a polynomial is 1 less than the degree of the polynomial. Here you have three local extrema. So the polynomial involved must be of even degree and greater than or equal to 4.

Fourth, there are an infinite number of polynomials of degree 4 that include the four given points. How might you find just one of them? But there is only one such quartic with a root of multiplicity 2 at one of the given points.

 

[Steven G]

If a polynomial crosses the x-axis at x=r, then one factor of the polynomial will be (x-r)^{an odd power}.
If a polynomial has a point that touches the x-axis at x=s, then one factor of the polynomial will be (x-s)^{an even power}.

What do you do with these factors to find a polynomial that crosses and touches the x-axis where you want them? How do you modify this polynomial so that it has the desired y-intercept?

 

[Otis]

Hi christie. We're told that the absolute maximum is unknown, so it cannot be one of the given points. In other words, the absolute maximum is not at the y-intercept; it must be somewhere in Quadrant I.



PS: To clarify in post #3, when Jomo talks about where a graph "touches the x-axis", he means a point where a graph reaches the x-axis but does not cross it.

?

 

[Steven G]

Hi christie. We're told that the absolute maximum is unknown, so it cannot be one of the given points. In other words, the absolute maximum is not at the y-intercept; it must be somewhere in Quadrant I.

View attachment 17237

PS: To clarify in post #3, when Jomo talks about where a graph "touches the x-axis", he means a point where a graph reaches the x-axis but does not cross it.

?

I am not sure if your comment about the max being in quadrant 1 is necessarily correct. Why can't the max be in quad 2?

This is just my opinion. Wherever you put your max then it is known. In your graph it is at (2,9). So why can't it be at the y-intercept. My logic is as follows. We were not given the point (0, 3), rather we read it off of the graph. I was not given the max but I still know it.

 

[topsquark]

I'm not clear on where the max should be. However it is possible to do this with a 4th degree polynomial and Otis's function gives a correct answer.

After toying with this for a while I can't find a higher degree polynomial that looks anything like the supplied graph.

-Dan

 

[Steven G]

The max can be in quad 1 or quad 2. In Otis' graph the function was increasing to the y intercept. There is no reason why it can't be decreasing to the y-intercept.

 

[Otis]

… In your graph [the absolute maximum] is at (2,9) …

Actually, it's at (√3, 4 + 8/3 · √3).

The [absolute maximum] can be in quad 1 or quad 2 …

Are you thinking of a higher-degree polynomial?

\(\;\)

 

[JeffM]

I'm not clear on where the max should be. However it is possible to do this with a 4th degree polynomial and Otis's function gives a correct answer.

After toying with this for a while I can't find a higher degree polynomial that looks anything like the supplied graph.

-Dan

There are an infinite number of polynomials of even degree that will satisfy what we believe was given. For example, all polynomials of the form, where n is a positive integer, work:

[MATH]f(x) = -\ \dfrac{1}{3^{(2n+2)}} * (x + 3)^{(2n + 2)}(x + 1)(x - 3)[/MATH]
clearly has exactly three x-intercepts, at minus 3, minus 1, and plus 3.

It has its y-intercept at [MATH]f(0) = - \ \dfrac{1}{3^{(2n+2)}} * 3^{(2n+2)} * 1 * (-\ 3) = 3.[/MATH]
Although an algebra student would not be expected to find this out,

[MATH]f'(x) = -\ \dfrac{1}{3^{(2n+2)}} * \{ (2n + 2)(x + 3)^{(2n+1)}(x + 1)(x - 3)(1) + (x + 3)^{(2n+2)}(1)(x - 3) + (x + 3)^{(2n+2)}(x + 1)(1)\} =[/MATH]
[MATH]- \dfrac{(x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{(2n + 1)(x + 1)(x - 3) + (x + 3)(x - 3) + (x + 3)(x + 1)\} =[/MATH]
[MATH]-\ \dfrac{(x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{(2n + 1)(x^2 - 2x - 3) + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]-\ \dfrac{(x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{(2n + 3)x^2 - (4n - 2)x - (6n + 9)\}.[/MATH]
If x < - 3, f'(x) is everywhere positive (check). If x > 3, f'(x) is everywhere negative (check). And f'(x) = 0 at
x = - 3 (check). But there are two other points where f'(x) = 0, namely where

[MATH](2n + 3)x^2 - (4n + 2)x - (6n + 9) = 0 \implies x = \dfrac{4n + 2 \pm \sqrt{16n^2 + 16n + 4 + 48n^2 + 144n + 108}}{4n + 6} \implies[/MATH]
[MATH]x = \dfrac{4n + 2 \pm \sqrt{60n^2 + 160n + 112}}{4n + 6} = \dfrac{2n + 1 \pm \sqrt{15n^2 + 40n + 28}}{2n + 3}.[/MATH]
If I did not mess up the algebra (which I may have done because I cannot validate otis's calculation of where the local extrema occur in the special case of n = 1), there will always be a local minimum in quadrant II and a local maximum in quadrant I.

 

[Steven G]

Actually, it's at (√3, 4 + 8/3 · √3).




Are you thinking of a higher-degree polynomial?

\(\;\)

No I am not thinking higher-degree polynomial. In your diagram imagine moving your max point to the left into quadrant 2, without disturbing the other key points given. Another words maybe the max occurs at x=-0.5 instead of at x~2

 

[Dr.Peterson]

No I am not thinking higher-degree polynomial. In your diagram imagine moving your max point to the left into quadrant 2, without disturbing the other key points given. Another words maybe the max occurs at x=-0.5 instead of at x~2

Since there is only one fourth-degree polynomial that has the indicated intercepts (and multiplicities), it would have to be a higher-order polynomial.

I think what you mean is that, given only what the student has learned, and not having determined the polynomial yet, it is conceivable, hypothetically, that the max is on either side of the y-axis. I agree.

Now, having lots of experience seeing polynomials, I can tell at a glance that it would have to bend the wrong way to do that; but the student can't make any such assumption. (That's why I mentioned "the odd wiggles".) Fortunately, it isn't needed.

 

[JeffM]

[MATH]f(x) = -\ \dfrac{1}{3^{(2n+2)}} * (x + 3)^{(2n+2)}(x + 1)(x - 3)[/MATH] is correct,

but I did make a mistake with respect to the derivative in my previous post. (Copied a term incorrectly.)

[MATH]f'(x) = -\ \dfrac{1}{3^{(2n+2)}} * \{ (2n + 2)(x + 3)^{(2n+1)}(x + 1)(x - 3)(1) + (x + 3)^{(2n+2)}(1)(x - 3) + (x + 3)^{(2n+2)}(x + 1)(1)\} =[/MATH]
[MATH] \dfrac{-\ (x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{(2n + 2)(x + 1)(x - 3) + (x + 3)(x - 3) + (x + 3)(x + 1)\} =[/MATH]
[MATH] \dfrac{-\ (x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{(2n + 2)(x^2 - 2x - 3) + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]-\ \dfrac{-\ (x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{2n(x^2 - 2x - 3) + 2(x^2 - 2x - 3) + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]\dfrac{-\ (x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{2nx^2 - 4nx - 6n + 2x^2 - 4x - 6 + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]\dfrac{-\ (x + 3)^{(2n+1)}}{3^{(2n+2)}} * \{(2n + 4)x^2 - 4nx - (6n + 12)\}.[/MATH]
So obviously f'(-3) = 0 as we want. And

[MATH]x < -\ 3 \implies (2n + 4)x^2 - 4nx - (6n + 12) >[/MATH]
[MATH]18n + 36 + 12n - 6n - 12 > 24n - 24 > 0 \ \because n > 0.[/MATH]
So if x < - 3, then f'(x) > 0, as we want.

There are only two other real points where f'(x) = 0, namely where

[MATH]x = \dfrac{4n \pm \sqrt{16n^2 - 4(2n + 4)\{-(6n + 12)\}}}{2(2n + 4)}[/MATH]
[MATH]\dfrac{4n \pm \sqrt{16n^2 + 4(12n^2 + 48n + 48)}}{2(2n + 4)} =[/MATH]
[MATH]\dfrac{4n \pm \sqrt{64n^2 + 192n + 192}}{2(2n + 4)} = \dfrac{4n \pm 8\sqrt{n^2 + 3n + 3}}{2(2n + 4)} =[/MATH]
[MATH]\dfrac{2n}{2n + 4} \pm \dfrac{4\sqrt{n^2 + 3n + 3}}{2n + 4}.[/MATH]
[MATH]x = \dfrac{2n}{2n + 4} + \dfrac{4\sqrt{n^2 + 3n + 3}}{2n + 4} \implies x > 0 \ \because n \ge 1.[/MATH]
So the local maximum must be in quadrant I.

[MATH]n^2 + 3n + 3 > n^2 \ \because n \ge 1.[/MATH]
[MATH]\therefore \dfrac{4\sqrt{n^2 + 3n + 3}}{2n + 4} > \dfrac{4n}{2n + 4} > \dfrac{2n}{24 + 4} \implies [/MATH]
[MATH]\therefore \dfrac{4\sqrt{n^2 + 3n + 3}}{2n + 4} - \dfrac{2n}{2n + 4} > 0 \implies [/MATH]
[MATH]\dfrac{2n}{24 + 4} - \dfrac{4\sqrt{n^2 + 3n + 3}}{2n + 4} < 0.[/MATH]
Thus, the local minimum must be in quadrant II.

I have some confidence in these results because I now agree with otis, who looked at the special case of n = 1. His function must be

[MATH]f(x) = -\ \dfrac{1}{3^{(2*1+2)}} * (x + 3)^{(2*1+2)}(x + 1)(x - 3) = -\ \dfrac{1}{81} * (x + 3)^4(x + 1)(x - 3)[/MATH]
which does satisfy the known points of the graph and agrees with my general solution.

[MATH]\therefore f'(x) = -\ \dfrac{1}{81} * \{4(x + 3)^3(x + 1)(x - 3) + (x + 3)^4(x - 3) + x^4(x + 1) =[/MATH]
[MATH]\dfrac{-\ (x + 3)^3}{81} * \{4(x + 1)(x - 3) + (x + 3)(x - 3) + (x + 3)(x + 1)\}[/MATH]
[MATH]\dfrac{-\ (x + 3)^3}{81} * \{4(x^2 - 2x - 3) + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]\dfrac{-\ (x + 3)^3}{81} * \{4(x^2 - 2x - 3) + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]\dfrac{-\ (x + 3)^3}{81} * \{4x^2 - 8x - 12 + x^2 - 9 + x^2 + 4x + 3\} = -\ \dfrac{-\ (x + 3)^3}{81} * (6x^2 - 4x - 18).[/MATH]
Notice that this agrees with my general formula 6 = 2 * 1 + 4, - 4 = - 4 * 1, and - 18 = - (6 * 1 + 12).

 

[JeffM]

Since there is only one fourth-degree polynomial that has the indicated intercepts (and multiplicities), it would have to be a higher-order polynomial.

I think what you mean is that, given only what the student has learned, and not having determined the polynomial yet, it is conceivable, hypothetically, that the max is on either side of the y-axis. I agree.

Now, having lots of experience seeing polynomials, I can tell at a glance that it would have to bend the wrong way to do that; but the student can't make any such assumption. (That's why I mentioned "the odd wiggles".) Fortunately, it isn't needed.

Please see my second post. Even a higher degree polynomial will not work. There are more constraints on this problem than superficially appear.

 

[Steven G]

Why is this not possible for a 4th degree polynomial?

 

[Dr.Peterson]

Please see my second post. Even a higher degree polynomial will not work. There are more constraints on this problem than superficially appear.

My point is only that this is far, far beyond the original problem, at least as it was surely intended. I think it was very poorly written, because it should explicitly say what is required, and should not show a contrary-to-fact graph, which can only confuse students. As I take it (having seen many problems at this level), it asks for a polynomial function either the intercepts and end behavior shown, and tells us to ignore where extrema appear to be. That's what it should say.

Feel free to deal with the more advanced question of whether any possible polynomial could have those intercepts and also have a maximum at x<0. I just don't think that's what the OP needs to hear; that's between you and Jomo (and possibly only yourself).

 

[Dr.Peterson]

View attachment 17252
Why is this not possible for a 4th degree polynomial?

Because, as I said before, the intercepts alone are sufficient to fully determine the unique fourth-degree polynomial, and it doesn't look like that.

I don't know that there is any a priori reason to make that conclusion, but it is a fact.

 

[Otis]

[MATH]f(x) = -\ \dfrac{1}{3^{(2n+2)}} * (x + 3)^{(2n+2)}(x + 1)(x - 3)[/MATH] …

[the results] agree with otis, who looked at the special case of n = 1 …

When n = 0, the expression above matches the quadratic polynomial whose graph I posted.

 

[JeffM]

View attachment 17252
Why is this not possible for a 4th degree polynomial?

I think I better stop posting. I have made errors in both my previous posts. Let's try for a THIRD time. Maybe I shall get it right this time.

We all agree that f(x) is to be a polynomial of even degree > 2 with exactly 3 roots, at -3, -1, and 3. Plus the root at - 3 must have multiplicity. Finally, f(0) = 3.

Thus, f(x) must have the form

[MATH]f(x) = a(x + 3)^{2n}(x + 1)(x - 3).[/MATH]
You with me so far?

To keep things neat, I shall solve for the non-zero constant a last, but note that a < 0.

[MATH]\therefore f'(x) = a\{(2n)(x + 3)^{(2n-1)}(x + 1)(x - 3)(1) + (x + 3)^{2n}(1)(x - 3) + (x + 3)^{2n}(x + 1)(1) =[/MATH]
[MATH]a(x + 3)^{(2n-1)}\{(2n(x + 1)(x - 3) + (x + 3)(x - 3) + (x + 3)(x + 1)\} =[/MATH]
[MATH]a(x + 3)^{(2n-1)}\{(2nx + 2n)(x - 3) + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]a(x + 3)^{(2n-1)}\{2nx^2 - 4nx - 6n + x^2 - 9 + x^2 + 4x + 3\} =[/MATH]
[MATH]a(x + 3)^{(2n-1)}\{(2n + 2)x^2 - 4(n - 1)x - (6n + 6)\}.[/MATH]
Let's hope I got the derivative correct this time.

[MATH]\therefore f'(0) = 0 \implies x = - 3 \text { or }[/MATH]
[MATH]x = \dfrac{4(n - 1) \pm \sqrt{16(n - 1)^2 - 4(2n + 2)(-1)(6n + 6)}}{2(2n + 2)} =[/MATH]
[MATH]\dfrac{4(n - 1) \pm \sqrt{16(n^2 - 2n + 1) + 4(12n^2 + 24n + 12)}}{2(2n + 2)} =[/MATH]
[MATH]\dfrac{4(n - 1) \pm \sqrt{16n^2 - 32n + 16 + 48n^2 + 96n + 48}}{2(2n + 2)} =[/MATH]
[MATH]\dfrac{4(n - 1) \pm \sqrt{64n^2 + 64n + 64}}{2(2n + 2)} = \dfrac{4(n - 1) \pm 8\sqrt{n^2 + n + 1}}{2(2n + 2)} =[/MATH]
[MATH]\dfrac{2(n - 1) \pm 4\sqrt{n^2 + n + 1}}{2n + 2}.[/MATH]
[MATH]\dfrac{2(n - 1) \+ 4\sqrt{n^2 + n + 1}}{2n + 2} > 0 \ \because n \ge 1.[/MATH]
Therefore, the local maximum must be in quadrant I, no matter what even degree the polynomial has.

[MATH](n - 1)^2 = n^2 - 2n + 1 < n^2 + n + 1 \implies 2(n - 1) < 2\sqrt{n^2 + n +1} < 4\sqrt{n^2 + n +1} \implies[/MATH]
[MATH]2(n - 1) - 4\sqrt{n^2 + n +1} < 0.[/MATH]
Therefore, Therefore, the local maximum must be in quadrant II, no matter what even degree the polynomial has.

As for a

[MATH]f(0) = 3 = a * 3^{2n} * 1 * (-\ 3) \implies a * 3^(2n) = -\ 1 \implies a = - \ \dfrac{1}{3^{2n}}.[/MATH]
If we take the special case of n = 1, the local minimum is when

[MATH]x = \dfrac{2(1 - 1) - 4\sqrt{1^2 + 1 + 1}}{2 * 1 + 2} = \dfrac{-\ 4\sqrt{3}}{4} = -\ \sqrt{3}.[/MATH]
The local maximum is when

[MATH]x = \dfrac{2(1 - 1) + 4\sqrt{1^2 + 1 + 1}}{2 * 1 + 2} = \sqrt{3}.[/MATH]
To sum up, there are an infinite number of polynomials of even degree, each with a local minimum in quadrant II and a local maximum in quadrant I. If I screwed this up again, I shall go to the corner for a day.