Question

11.5.3 quiz: graphs of trigonometric functions
which graph shows an odd function?
(images of graphs labeled a, b, c, d with corresponding options)

Explanation:

Step1: Recall odd function property

An odd function satisfies \( f(-x) = -f(x) \), so its graph is symmetric about the origin.

Step2: Analyze each graph

• Graph A: Symmetric about the y - axis (even function property), not odd.
• Graph B: Let's check symmetry. For an odd function, if we reflect over the origin, the graph should map onto itself. Graph B: When we consider \( x \) and \( -x \), the shape shows symmetry about the origin? Wait, no, let's re - check. Wait, Graph B: Let's see the behavior. Wait, actually, Graph D: Wait, no, let's look again. Wait, the key is origin symmetry.
• Graph C: Not a trigonometric graph in the standard sense, and not symmetric about origin.
• Graph D: The graph passes through the origin, and if we rotate it 180 degrees about the origin, it maps onto itself. Also, for a trigonometric function, sine function is odd, and its graph (like \( y = \sin x \)) is odd. Graph D resembles the sine - like graph with origin symmetry. Also, Graph B: Wait, no, let's correct. Wait, Graph A is cosine - like (even), Graph B: Let's check the y - intercept. Graph B has a y - intercept at negative, but when we check \( f(-x) \), for an odd function, \( f(0)=0 \) (since \( f(0)=-f(0)\implies f(0) = 0 \)). Graph A has \( f(0)=1 \), Graph B has \( f(0)

eq0 \), Graph D has \( f(0) = 0 \). Also, the symmetry: Graph D is symmetric about the origin. So Graph D shows an odd function. Wait, but in the options, let's re - evaluate. Wait, the correct graph for an odd function (trigonometric) should have origin symmetry. Among the options, Graph D (the last one) has origin symmetry, and also, the function \( y=\sin x \) is odd, and its graph passes through the origin and is symmetric about the origin. Also, Graph B: No, Graph B's y - intercept is not zero. Graph A's y - intercept is 1 (even function). Graph D: passes through (0,0) and is symmetric about origin. So the graph that shows an odd function is Graph D? Wait, no, wait the options: Wait, the user's graph labels: A, B, C, D. Wait, maybe I made a mistake. Wait, let's re - check the graphs:

Wait, the first graph (A) is like \( y=\cos x \) (even), second (B) is a graph that is not symmetric about origin? Wait, no, maybe the correct one is Graph B? No, wait, no. Wait, the key is that for an odd function, \( f(0) = 0 \) (because \( f(0)=-f(0)\implies 2f(0)=0\implies f(0) = 0 \)). So graphs with \( f(0)
eq0 \) (like A and B) can't be odd. Graph D has \( f(0) = 0 \), and is symmetric about origin. Also, the graph of an odd trigonometric function (like sine) has origin symmetry. So the correct graph is D? Wait, but maybe I messed up. Wait, let's check again.

Wait, the problem is about trigonometric functions. The odd trigonometric functions (like \( y = \sin x \)) have graphs symmetric about the origin. So the graph that is symmetric about the origin is the one for the odd function. Among the options, the graph labeled D (the last one) has origin symmetry and \( f(0)=0 \), so it represents an odd function.

Answer:

D (the graph labeled D, the last one in the given set of graphs)