Construct A Polynomial Function

[mathdad]

Construct a polynomial function given a graph that crosses the x-axis at x = -1, x = 1, and x = 2 and crosses the y-axis at the point
(0, 1).

Note: The answer given is
f(x) = -(1/2)(x + 1)(x - 1)^2(x - 2).

Questions:

A. Where did -1/2 come from?

B. Where did the power of 2 for the factor
(x - 1)^2 come from?

C. Can I use f(x) = a(x - r)(x - r)(x - r)(x - r) to construct the polynomial function given here as the answer?

D. The degree for the expanded form of this function is 4. Yes?

 

[Romsek]

If I was given the info given in the problem I'd proceed as

\(\displaystyle p(x) = a(x+1)(x-1)(x-2)\)

\(\displaystyle p(0)=1\\ a(1)(-1)(-2) = 1\\a = \dfrac 1 2\\p(x) = \dfrac 1 2 (x+1)(x-1)(x-2)\)

I can't say why they chose to make the \(\displaystyle (x-1)\) term a square. It's not necessary.
Any chance the problem specifies a degree four polynomial?

The problem does say "construct a polynomial..." it doesn't say construct the minimum degree polynomial.
Still, their answer seems odd without further explanation as to why they chose what they did.

Yes. Their answer is a degree 4 polynomial.

 

[mathdad]

If I was given the info given in the problem I'd proceed as

\(\displaystyle p(x) = a(x+1)(x-1)(x-2)\)

\(\displaystyle p(0)=1\\ a(1)(-1)(-2) = 1\\a = \dfrac 1 2\\p(x) = \dfrac 1 2 (x+1)(x-1)(x-2)\)

I can't say why they chose to make the \(\displaystyle (x-1)\) term a square. It's not necessary.
Any chance the problem specifies a degree four polynomial?

The problem does say "construct a polynomial..." it doesn't say construct the minimum degree polynomial.
Still, their answer seems odd without further explanation as to why they chose what they did.

Yes. Their answer is a degree 4 polynomial.

The problem does not specify a degree of 4. The problem goes on to say that answers will vary but no explanation as to why answers will vary. Your value for a is 1/2. The value for a in the textbook answer is -1/2.

 

[Romsek]

the square makes the (x-1) term positive at 0 whereas without it it's negative. Hence the difference in the signs of the leading coefficient.

 

[mathdad]

the square makes the (x-1) term positive at 0 whereas without it it's negative. Hence the difference in the signs of the leading coefficient.

Good to know.

 

[Dr.Peterson]

Construct a polynomial function given a graph that crosses the x-axis at x = -1, x = 1, and x = 2 and crosses the y-axis at the point
(0, 1).

Note: The answer given is f(x) = -(1/2)(x + 1)(x - 1)^2(x - 2).

B. Where did the power of 2 for the factor (x - 1)^2 come from?

The problem does not specify a degree of 4. The problem goes on to say that answers will vary but no explanation as to why answers will vary.

It's nice to know that answers are expected to vary (as they should); but their answer doesn't fit the problem, since the graph doesn't cross the x-axis at x = 1, but instead "touches and turns" because of the squaring of (x-1). So unless you copied something wrong, their answer is not valid.

For a valid answer, you can have any odd exponent on any factor(s), and you could include other factors as long as it isn't saying there are no other x-intercepts. The multiplier must then be determined based on the particular factors used.

 

[mathdad]

It's nice to know that answers are expected to vary (as they should); but their answer doesn't fit the problem, since the graph doesn't cross the x-axis at x = 1, but instead "touches and turns" because of the squaring of (x-1). So unless you copied something wrong, their answer is not valid.

For a valid answer, you can have any odd exponent on any factor(s), and you could include other factors as long as it isn't saying there are no other x-intercepts. The multiplier must then be determined based on the particular factors used.

Perhaps it is a typo at my end. Either way, interesting conversation. Just having fun with Sullivan's textbook.

 

[Steven G]

If I was given the info given in the problem I'd proceed as

\(\displaystyle p(x) = a(x+1)(x-1)(x-2)\)

\(\displaystyle p(0)=1\\ a(1)(-1)(-2) = 1\\a = \dfrac 1 2\\p(x) = \dfrac 1 2 (x+1)(x-1)(x-2)\)

I can't say why they chose to make the \(\displaystyle (x-1)\) term a square. It's not necessary.
Any chance the problem specifies a degree four polynomial?

The problem does say "construct a polynomial..." it doesn't say construct the minimum degree polynomial.
Still, their answer seems odd without further explanation as to why they chose what they did.

Yes. Their answer is a degree 4 polynomial.

Professor, you are missing one point. The problem said that the graph crosses the x-axes at x=1, so the factor (x-1) can not be raised to an even power.

 

[mathdad]

Professor, you are missing one point. The problem said that the graph crosses the x-axes at x=1, so the factor (x-1) can not be raised to an even power.

Good to know.