Impossible pre cal question Give formula for the graph of the polynominal

[xdrgnh]

Homework Statement

http://www.math.poly.edu/courses/ma0914/past_exams/MA0922_Final_2000-12-13.pdf First problem on the exam

Homework Equations

I believe the answer should be a 3rd degree polynomial so Ax^3+Bx^2+Cx+D

The Attempt at a Solution

Ax^3+Cx
-8A-2C=0
8A+2C=0

-A-C=3
-4A=C
-A+4A=3A

A=1.

However that is impossible because A has to be negative for the 2nd derivative on the left side of the equation to be positive. Idk where I went wrong.

 

[DeIdeal]

I believe the answer should be a 3rd degree polynomial so Ax^3+Bx^2+Cx+D

This is not correct. It should be a 5th degree polynomial. Are you able to explain why & continue from there?

There are several possible ways to form the necessary equations to find out the constants.

 

[xdrgnh]

I guess I would continue by plugging in the values for the zeros and plug in the values for (-1,3) and (1,-3) and solve the system of linear equations. However I don't know why it's a 5 degree. Technically I'm not supposed to use derivatives.

 

[DeIdeal]

I guess I would continue by plugging in the values for the zeros and plug in the values for (-1,3) and (1,-3) and solve the system of linear equations. However I don't know why it's a 5 degree.

Something like that, yeah : ]

I'm not too familiar with the US (I assume?) education system, so I don't know what is taught in Pre-calculus. If, however, you already know how to take the derivative of polynomials, you might want to use that in order to make the system of equations easier to solve. It shouldn't make a huge difference, though.

As for why it's 5th degree: You should be able to see it from the shape of the graph. One way to explain it is that a 3rd degree polynomial can only change direction two times (ie. f'(x) has only two roots and is a 2nd degree polynomial). You can see that the graph changes from increasing to decreasing or vice versa four times, so f'(x) has four roots and f(x) is a 5th-degree polynomial. Well, at least that's one possibility, and you were only looking for ONE possible formula.

 

[xdrgnh]

Alright I got it. I Went from Algebra 2 and trig to calc 1 and 2 and then all of the way up to complex variables. However to get a job as a math tutor I need to take a test in pre calculus. However contrary to the name pre calculus I noticed I never used any of this stuff in my upper math classes.

 

[xdrgnh]

-32A+16B-8C+4D-2E=0
32A+16B+8C+4D+2E=0
-A+B-C+D-E=3
A+B+C+D=-3

So I get 4 equations and 5 unknowns. However when I try to solve them I get.
2B+2D=0
32B+8D=0
which implies that both B and D are zero does that seem right so far.

edit my bad.

 

[DeIdeal]

-32A+16B-8C+4D-2E=0
32A+16B+8C+4D+2E=0
-A+B-C+D-E=3
A+B+C+D=-3

So I get 4 equations and 4 unknowns. However when I try to solve them I get.
2B+2D=0
32B+8D=0
which implies that both B and D are zero does that seem right so far.

Yeah, B=0, D=0.

EDIT: You should get 5 equations for the 5 unknowns, though.

Well, six for the six unknowns, but f(0)=0 -> F=0 is pretty damn obvious, so.

 

[xdrgnh]

Where would the 5th one come from besides the face that F would equal zero. Because I'm getting something that can have infinitely many solutions.

 

[DeIdeal]

Where would the 5th one come from besides the face that F would equal zero. Because I'm getting something that can have infinitely many solutions.

You can use f'(2)=0 or f'(-2)=0, for example. There's an exact solution, don't worry.

 

[xdrgnh]

I can't use derivatives this pre calc. Is this question possible without calculus?

 

[xdrgnh]

Also wouldn't the 2nd derivative be zero not the first at 2. After looking I noticed that the 1st derivative is zero also. Not like it helps.

 

[DeIdeal]

Oh. To be honest, I didn't really think about that possibility. Using derivatives just seemed so obvious that I didn't bother to check how to do it without them, sorry.

I'm not sure about this, but I don't see any way to do this without the derivatives.

EDIT: No, the first derivative is zero at the (local) maxima and minima. The sign of the second derivative tells whether it's a maximum or a minimum, unless f''(x)=0.

 

[xdrgnh]

Well even though my original hunch that it's not possible is correct I see that my reasoning was wrong. Thanks for the help I learned something.

 

[Filip Larsen]

It is a bit easier if you are familiar with the fundamental theorem of algebra, from which it follows that if a n-degree polynomial has the roots (or zeros) x₁, x₂, ..., x_n (some of which may be equal to each other), then the polynomial can be written as g(x) = a(x-x₁)(x-x₂)...(x-x_n), where a root will be repeated if the function has zero derivative (i.e. horizontal tangent) in that root.

Looking at the graph you can count 3 roots, of which 2 is repeated since they have a horizontal tangent, and 4 horizontal tangents in total. Taken together that means g must be a 5-degree polynomial with the root in -2 and 2 each repeated twice. From the last information, that g(-1) = 3, you can then solve for a and get one particular polynomial that "fits" all the information given.

 

[xdrgnh]

I wrote down the g(-1)=3 and got 4 equations with 5 unknowns.

 

[D H]

The graph appears to be <fill in the blank> about x=0. Use that.

 

[xdrgnh]

symmetric?

 

[D H]

Close. There is a symmetry there.

Hint: How do f(x) and f(-x) appear to be related?

 

[klondike]

f(-x)=-f(x) which allow us setting all even degree (include f) to zero. Makes life easier but still need the knowledge of derivative. How this can be done without derivative?

symmetric?

 

[skiller]

Filip Larsen (post #14) has suggested all you need to know to solve it. There's no need to bother about whether it's "symmetric" or looking at any derivatives or any simultaneous equations (as the first few posts were suggesting).

 

[D H]

Filip Larsen (post #14) has suggested all you need to know to solve it. There's no need to bother about whether it's "symmetric"

Well there is a need for symmetry considerations if you read the zero at x = -2 as a zero at "?" In other words, specifying where that zero occurs is part of the problem.

OTOH, if one reads the graph correctly, then post #14 says it all. No calculus needed.

 

[skiller]

EDIT:
**I've removed my question as I misread your post**

:smilie:

 

[Curious3141]

Homework Statement

http://www.math.poly.edu/courses/ma0914/past_exams/MA0922_Final_2000-12-13.pdf First problem on the exam

Homework Equations

I believe the answer should be a 3rd degree polynomial so Ax^3+Bx^2+Cx+D

The Attempt at a Solution

Ax^3+Cx
-8A-2C=0
8A+2C=0

-A-C=3
-4A=C
-A+4A=3A

A=1.

However that is impossible because A has to be negative for the 2nd derivative on the left side of the equation to be positive. Idk where I went wrong.

This question can be solved by just observation, and minimal calculation.

Whenever a polynomial curve "bounces" tangentially off the x-axis, it means there's a repeated root there. So there are repeated roots at x = -2 and x = 2. Meaning that (x-2)^2 and (x+2)^2 are both factors of the polynomial.

The curve also passes through the origin, meaning that f(0) = 0. Hence x is also a factor.

So you know f(x) can be of the form kx(x-2)²(x+2)²

You're also given f(-1) = 3, so put that into the above equation to work out k.

 

[SammyS]

symmetric?

It looks as though there is a lot of traffic on this thread, with an occasional visit by OP.

 

[xdrgnh]

Well guys I believe I got it via following post 14 advice. -(1 / 3) * x * (x + 2) ^ 2 * (x – 2) ^ 2. I remember learning about the fundamental theorem of Algebra in my junior year of high school but because I never used it in my physics and future math classes I forgot it. I got my pre caluclus test this Thursday and this Wednesday I'm going to be studying for it. Do you guys have good online resources. The problem with these test questions is that they have no answers to them.

 

[HallsofIvy]

Well even though my original hunch that it's not possible is correct I see that my reasoning was wrong. Thanks for the help I learned something.

No, it possible without Calculus if you recognise that x= 2 and x= -2 must be double roots. The equation must be of the form [itex]y= a(x- 2)^2(x+ 2)^2(x- 0) and the fact that y= 3 when x= -1 tells you that 3= a(9)(1)(-1) so a= -1/3.

 

[amiras]

The possible one is: -x(x^2-4)^2*1/3

If you are not allowed (or do not know yet) properties of derivatives you need to do some trial and error.

It is shown that this graph intersects with the x-axis at three points -2, 0 and 2. That means that the equation must be at least of polynomial 3. And should not have any more roots in the interval [-3, 3], Let's assume that it does not have any more roots.

So basically the required equation, should be in the form of C*(-x)(x-2)(x+2)=C(-x)(x^2-4) in order for the three roots to hold, where C is some constant that is only required to adjust for point (-1, 3). Now if you carefully check for several points this is not the right function. So the real one could be a multiplier of (x^2 -4). It cannot be a multiplier of x because it would bake a function even. So C(-x)(x^2-4)*(x^2 -4)=C(-x)(x^2-4)^2 works nicely.

 

[coolul007]

x^3 - 4x is the equation of the curve...

 

[xdrgnh]

Did you read the thread? That was my first guess also btw but it was wrong.

 

[coolul007]

Did you read the thread? That was my first guess also btw but it was wrong.

I thought I read the thread, must have missed something? Ok, I see the slope on the negative side is reversed, yes must be a 5th degree.