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enter image description here

Given the graph I'm asked for the degree.
I can see how it is odd. I can see how the $a$ value is negative. I can see the inflection point at $0$. My question is how would I know if this is degree 3 or 5 or 7....? The book's answer is this is degree 5.

It also says it starts in quadrant 1, and from my trig background, I'd say it starts (left to right) in quadrant 2.

But my main question is how is it degree 5 vs degree 3 or degree 7?

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    It is also not particularly clear whether it is in fact odd or not. We can't see what happens way far to the left. It is *implied* that it is odd, as what is pictured is usually the "important" details of the function, but it could for all we know turn around again at $x=-1000$ or something. – 2017-02-20

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Just to reiterate the comments made above, the question itself is flawed. An image of only a finite section of the graph of a function is not enough information to conclude much of anything except for lower bounds on degrees.

Here are a few graphs which look very similar to the original graph in the question. Note, these are all graphs of different functions! I did not just copy the same image three times to make a point!

$$y=-\frac{x^5}{20}+\frac{2x^3}{3}$$

degree five

This is probably the function used to generate the image in the textbook, however look at these others...

$$y=\frac{x^7}{10^{10}}-\frac{x^5}{20}+\frac{2x^3}{3}$$

degree seven

Ironically, this graph will have the opposite horizontal limits as the original graph. In fact, as $x\to-\infty$ this graph will become very negative! not positive! This is despite what we can see from the image.

One more:

$$y=\frac{x^6}{10^{10}}-\frac{x^5}{20}+\frac{2x^3}{3}$$

degree six

Here in this picture, the polynomial isn't even of odd degree, but rather is even.

What we can determine is that since the number of inflection points is visibly greater than or equal to $3$, the degree of the polynomial must be at a minimum $5$ but could be more than that too. We know this because inflection points occur when the second derivative is equal to zero. As such we know that the second derivative of the polynomial in question is degree at least three. By integrating twice we get the original function must be a polynomial of degree at least five.

Any further information beyond this cannot be gleaned from the picture alone. The correct answer to the originally phrased question is:

  • Is the function even or odd: unknown (but appears odd from the given information as any image of a function should include as much relevant information as possible)

  • What is the degree of the function: at least five (it appears to be five but it could be more)

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    Great explanation. Fully clarified for me now. – 2017-02-20
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Count the number of inflection points. That rules out degree $3$ (a third degree polynomial can only have one inflection point). As for degree $7$, there is no way to really tell from the graph that there isn't a $10^{-100}x^7$ term in there (or a $10^{-100}x^6$ for that matter, which would ruin the odd-ness), so if it truly is degree $5$, then that's not something we can see from a picture like this.

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    Is the number of inflection points = 3? One at the local min, one at the local max and one at zero? Is the movement of the graph at zero considered an inflection point because it changes in concavity? – 2017-02-20
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    @user163862 inflection points are where it changes concavity from concave up to concave down or vice versa. You have *at the minimum* the three visible inflection points, one near $x=-2$, one near $x=0$, and another near $x=2$. There could be more outside of the visible window of the graph, just as these inflection points could be repeated. Still, if we were told that this is the graph of a polynomial, since the degree is always strictly greater than the number of inflection points (*with the exception of constant polynomial*) it could not possibly be the graph of a degree three polynomial. – 2017-02-20
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You can see two roots at approximately $x = 4$ and $x = -4$. Plus, there is a root of multiplicity $3$ at $x = 0$. Therefore it is an equation of degree $5$.

It's actually not possible to determine exactly what the degree of the equation actually is from a graph. But $5$ is the answer they're looking for.

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    How can you tell that the root at $x=0$ has a multiplicity of 3? – 2017-02-20
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    Can I determine the degree based on the number of extrema points? – 2017-02-20
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    @user163862 you can tell a lower bound to the degree based on the number of local extrema, but not explicitly the degree itself. – 2017-02-20
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    @user163862 because that root "looks like" a cubic. Mind you, even that isn't perfect because it could "look like" a quintic ect. But that's my logic. – 2017-02-20