the function ( f(x) ) is graphed below. determine whether the degree of the function is even or odd and…

Explanation:

Step1: Recall the end - behavior of even and odd degree functions

For a polynomial function (y = f(x)), if the degree (n) is even:

• When (n>0), as (x\to\pm\infty), if the leading coefficient (a_n>0), (y = f(x)\to+\infty); if (a_n < 0), (y=f(x)\to-\infty) (both ends of the graph go in the same direction). If the degree (n) is odd:
• As (x\to+\infty) and (x\to-\infty), (y = f(x)) go in opposite directions. Looking at the graph of (y = f(x)), as (x\to-\infty), (y = f(x)\to+\infty) and as (x\to+\infty), (y = f(x)\to-\infty). The ends of the graph go in opposite directions, so the degree of the function is odd.

Step2: Recall the symmetry of even and odd functions

An even function (f(x)) satisfies (f(-x)=f(x)), and its graph is symmetric about the (y) - axis. An odd function (f(x)) satisfies (f(-x)=-f(x)), and its graph is symmetric about the origin. Take a point ((x,y)) on the graph. For example, if we assume a non - zero (x) value (say (x = 30), (y\approx0)). If we consider (-x=-30), the (y) - values for (x = 30) and (x=-30) are not such that (y(-30)=y(30)) (symmetry about (y) - axis for even functions) or (y(-30)=-y(30)) (symmetry about the origin for odd functions). We can also use the general property: if (f(x)) is a polynomial (f(x)=a_nx^n+\cdots+a_1x + a_0), for an odd function (a_{2k}=0) ((k = 0,1,\cdots)) and for an even function (a_{2k + 1}=0) ((k=0,1,\cdots)). Since the degree is odd (from Step 1), but the graph is not symmetric about the origin (tested by the symmetry property), the function has an odd degree but is not an odd function.

Answer:

(f(x)) has an odd degree, but is not an odd function.