Working backwards to create a Polynomial

GuavaEater

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Feb 9, 2017
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Hey there,

I'm having trouble with this question from my polynomials/graphing unit.

1. Write the equation of the polynomial function in factored form that has the following properties:
- Is tangent to the x-axis at (-2,0)
- 1/3 is a root with a multiplicity of 2
- 1 is a root with a multiplicity of 3
- has a y-intercept of 2/3

What I tried doing was writing it first in non-factored form first, which I'm not even sure is correct. It looked like this:

f(x) = x3 +(1/3)x2-2x+(2/3)

My rationale was that -2, 1/3, and 1 are all x-intercepts, and 2/3 is the y-intercept. I then attempted synthetic division using 1 as the restrictiion which did not yield any results. How would I go about creating this factored function?
 
Hey there,

I'm having trouble with this question from my polynomials/graphing unit.

1. Write the equation of the polynomial function in factored form that has the following properties:
- Is tangent to the x-axis at (-2,0)
- 1/3 is a root with a multiplicity of 2
- 1 is a root with a multiplicity of 3
- has a y-intercept of 2/3

What I tried doing was writing it first in non-factored form first, which I'm not even sure is correct. It looked like this:

f(x) = x3 +(1/3)x2-2x+(2/3)

My rationale was that -2, 1/3, and 1 are all x-intercepts, and 2/3 is the y-intercept. I then attempted synthetic division using 1 as the restrictiion which did not yield any results. How would I go about creating this factored function?
You work your way up from factored form.

A function with a root at x = a with a multiplicity of 'n' - will have (x-a)n as a factor.

What does your strating function look like with that knowledge and:

- 1/3 is a root with a multiplicity of 2
- 1 is a root with a multiplicity of 3

Please show your work starting here.
 
I'm having trouble with this question from my polynomials/graphing unit.

1. Write the equation of the polynomial function in factored form that has the following properties:
- Is tangent to the x-axis at (-2,0)
- 1/3 is a root with a multiplicity of 2
- 1 is a root with a multiplicity of 3
- has a y-intercept of 2/3

What I tried doing was writing it first in non-factored form first, which I'm not even sure is correct. It looked like this:

f(x) = x3 +(1/3)x2-2x+(2/3)

My rationale was that -2, 1/3, and 1 are all x-intercepts, and 2/3 is the y-intercept.
Whoever told you that zeroes are found by considering the coefficients of terms was very badly mistaken. :shock: Instead, one finds zeroes from factors. For instance, y = 2x - 6 has a zero at x = 3, thought 3 is neither of the coefficients.

Do you know what "with a multiplicity of (number)" means? Do you know what "roots" are, especially as they relate to x-intercepts and zeroes? How has your class been taught to deal with points of tangency?

Please be complete. Thank you! ;)
 
Please read the previous two comments carefully.

I hope that you did not write the problem exactly as it is given in your book because there is no unique answer to that problem.

It would be better phrased as find the "equation of the polynomial of lowest degree that has ..."
 
So I realized my error, and what I believe the answer to be is f(x)= (x-1/3)2(x-1)3(x+2)

Each of those are the roots, and that is in factored form. Lastly, the multiplicity refers to the degree of each term, (2,3,1).

Thanks for the help guys!
 
So I realized my error, and what I believe the answer to be is f(x)= (x-1/3)2(x-1)3(x+2)

Each of those are the roots, and that is in factored form. Lastly, the multiplicity refers to the degree of each term, (2,3,1).

Thanks for the help guys!
Your function is "almost" correct. It does NOT satisfy the following conditions:

- Is tangent to the x-axis at (-2,0)

- has a y-intercept of 2/3
 
Try substituting x = 0 into \(\displaystyle \left ( x - \dfrac{1}{3} \right )^2(x - 1)^3(x + 2)\).

Do you get 2/3?

You were asked earlier what you knew about points of tangency of a polynomial to a horizontal line, particularly, the x-axis. It is very hard for us to answer you when we do not know what you know. Do you know any differential calculus?
 
I'm just learning what tangency is, and this is a grade 12 regular stream math course, so no calc knowledge.
 
I'm just learning what tangency is, and this is a grade 12 regular stream math course, so no calc knowledge.
Then I would guess that they're possibly referring to x-intercepts that don't "intersect" so much as "just touch", being those of even degree. So throw a couple more factors in there.... ;)
 
I'm just learning what tangency is, and this is a grade 12 regular stream math course, so no calc knowledge.
Once you have a polynomial with the right points of tangency, you then need to make sure that the polynomial has a value of 2/3 if x = 0.

Couple of things to know.

(1) Every polynomial of degree 2n + 1 with real coefficients can be factored into 1 linear term with real coefficients and n quadratic terms with real terms. (It may or may not be possible to factor one or more of the quadratics into two linear terms with real coefficients.)

(2) Every polynomial of degree 2n with real coefficients can be factored into n quadratic terms with real coefficients. (It may or may not be possible to factor further.)

(3) If and only if \(\displaystyle (ax + b)^n\) is a factor of a polynomial and

\(\displaystyle n \in \mathbb Z,\ n > 0,\ and\ a \ne 0\),

then and only then is \(\displaystyle -\ \dfrac{b}{a}\) an x-intercept of that polynomial.

(4) If and only if \(\displaystyle (ax + b)^n\) is a factor of a polynomial and

\(\displaystyle n \in \mathbb Z,\ n > 1,\ and\ a \ne 0\),

then and only then is \(\displaystyle -\ \dfrac{b}{a}\) also a point of tangency to the x-axis.

It really is not fair to give this problem without telling you those four facts about polynomials.
 
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