# Question Video: Determining the Parity of a Graphed Rational Function Mathematics • 12th Grade

Is the function represented by the figure even, odd or neither even nor odd?

03:57

### Video Transcript

Is the function represented by the figure even, odd, or neither even nor odd?

This is a question about the parity of the function in the graph, and there are three options here. The function is either even, or itβs odd, or itβs neither. Letβs remind ourselves what the terms even and odd mean in the context of functions.

A function π of π₯ is even if π of minus π₯ equals π of π₯. That must hold for all values of π₯. So π of negative one must be π of one, π of negative seven must be π of seven, π of negative π must be π of π, and so on.

The definition of an odd function is very similar: the function π of π₯ is odd if π of minus π₯ equals minus π of π₯. So for an odd function, π of negative one is minus π of one, π of negative seven is minus π of seven, and π of negative π is minus π of π.

And unlike for whole numbers, our function π of π₯ doesnβt have to be either even or odd; it can be neither. Letβs first check if the function is even. In other words, is π of minus π₯ equal to π of π₯ for all values of π₯? Well, letβs try a value of π₯, say one, and find π of negative one.

So we look at negative one on the π₯-axis, we go up until we hit the graph, and we see that π of negative one is one. So for π of π₯ to be even, π of one has to be equal to one as well. We go through the same process to find π of one, going down this time until we hit the graph. And reading off from the π¦-axis, we conclude that π of one is equal to negative one. So clearly, π of negative one is not equal to π of one, and hence π of π₯ is not even.

Remember this doesnβt mean that π of π₯ is odd; π of π₯ could be neither even nor odd. We now check if π of π₯ is odd, so is π of minus π₯ equal to minus π of π₯ for all values of π₯? We can use the values of the function that we found already β itβs a shame to waste them β and we can see that π of negative one is equal to minus π of one.

On the left-hand side, we have one, and on the right-hand side, we have minus negative one, which is also equal to one. And so itβs true that π of negative π₯ is equal to minus π of π₯ when π₯ is one at least, but itβs not enough for π of minus π₯ to be equal to minus π of π₯ for one value of π₯; it has to be true for all values of π₯.

Checking another value of π₯, letβs say seven. While itβs hard to read off the values of π of negative seven and π of seven, itβs very plausible that their opposites π of negative seven is just a bit greater than zero and π of seven is just a bit less than zero. And we could continue trying values of π₯ until weβve convinced ourselves that π of π₯ is indeed an odd function.

Here weβve used the definition of an odd function to show that our function is indeed odd, but thereβs another way to do it; we can use a property of odd functions. A function is odd if its graph has a 180-degree rotational symmetry about the origin.

We can see that this is true for our graph: if we rotate our graph by 180 degrees about the origin, we get the same graph back. And thereβs a corresponding fact for an even function: a function is even if its graph is symmetric in the π¦-axis. So if you reflect the graph of an even function in the π¦-axis, you get the same graph back.

Looking at our graph, we can see that our graph isnβt symmetric in the π¦-axis, and so our function isnβt even, but as stated before, because our graph has a 180-degree rotational symmetry about the origin, our function is odd.