Graph of Polynomial Function

[mathdad]

A. Given a graph of a function, what are the traits that easily display it is a polynomial function?

B. How do you construct a polynomial function from a given polynomial graph of f?

NOTE: I seek solution steps only. I want to try this on my own (following your steps) before posting my work here.

 

[Dr.Peterson]

To answer A, I would look up what your book says about the appearance of a polynomial graph. They probably mention things like continuity and end behavior. (You are only asking for steps to find the answer, so I don't know what more I can say!)

Ultimately, though, you can never be sure that the graph represents a polynomial function; it is easy to draw something that looks generally like a polynomial, but that I can recognize from very subtle features is (probably) not exactly a polynomial. A slight wiggle in the wrong place can mess it all up.

I don't know that there is an answer to B! It depends on exactly what information is given in the graph (e.g. what points are known exactly), and whether you are allowed to make assumptions about the apparent degree. (There was a question recently where that issue led to some discussion.)

These are odd questions to see in a textbook or similar source! I don't expect them to ask for the impossible, or to tacitly expect you to make guesses.

 

[Otis]

For part A, I would start by reading about polynomial graphs and their attributes.

Step 1: Google keywords identifying polynomial graphs

Step 2: Study the information at several sites

?

 

[Otis]

For part B, you could google keywords like identifying polynomials from their graphs and study some examples.

However, not every method you see may work with any polynomial graph. For example, if I were given a polynomial graph which showed x-intercepts -4, -1, 3 and 7, then I would write

y = A(x + 4)(x + 1)(x - 3)(x - 7)

followed by substituting (x,y) coordinates from a fifth point, in order to solve for the leading coefficient A.

But, if I were given the graph of a parabola that doesn't cross the x-axis, then I would use the vertex coordinates (h,k) to write

y = A(x - h)^2 + k

followed by substituting (x,y) coordinates from another point, in order to solve for the leading coefficient A.

If the vertex coordinates were not shown, then I would use three known (x,y) points and the standard form to create and solve a system of three equations, in order to find the coefficients A,B,C.

y = Ax^2 + Bx + C

In other words, there is no single answer for part B. Methods will vary, depending on the information given.

?

 

[mathdad]

To answer A, I would look up what your book says about the appearance of a polynomial graph. They probably mention things like continuity and end behavior. (You are only asking for steps to find the answer, so I don't know what more I can say!)

Ultimately, though, you can never be sure that the graph represents a polynomial function; it is easy to draw something that looks generally like a polynomial, but that I can recognize from very subtle features is (probably) not exactly a polynomial. A slight wiggle in the wrong place can mess it all up.

I don't know that there is an answer to B! It depends on exactly what information is given in the graph (e.g. what points are known exactly), and whether you are allowed to make assumptions about the apparent degree. (There was a question recently where that issue led to some discussion.)

These are odd questions to see in a textbook or similar source! I don't expect them to ask for the impossible, or to tacitly expect you to make guesses.

I will post what the textbook has to say about the solution process.

 

[mathdad]

For part B, you could google keywords like identifying polynomials from their graphs and study some examples.

However, not every method you see may work with any polynomial graph. For example, if I were given a polynomial graph which showed x-intercepts -4, -1, 3 and 7, then I would write

y = A(x + 4)(x + 1)(x - 3)(x - 7)

followed by substituting (x,y) coordinates from a fifth point, in order to solve for the leading coefficient A.

But, if I were given the graph of a parabola that doesn't cross the x-axis, then I would use the vertex coordinates (h,k) to write

y = A(x - h)^2 + k

followed by substituting (x,y) coordinates from another point, in order to solve for the leading coefficient A.

If the vertex coordinates were not shown, then I would use three known (x,y) points and the standard form to create and solve a system of three equations, in order to find the coefficients A,B,C.

y = Ax^2 + Bx + C

In other words, there is no single answer for part B. Methods will vary, depending on the information given.

?

Wow! Another detailed reply. Thank you very much.