Question

which function is graphed below?
options:
$f(x) = \cos(x)$
$f(x) = -\cos(x)$
$f(x) = \sin(x)$
$f(x) = -\sin(x)$

Explanation:

Step1: Recall trigonometric graphs

The standard graphs of \(y = \cos(x)\), \(y=-\cos(x)\), \(y = \sin(x)\), \(y=-\sin(x)\) have distinct features. The graph of \(y=\sin(x)\) passes through the origin \((0,0)\), has a period of \(2\pi\), and at \(x = 0\), \(\sin(0)=0\). The graph of \(y=\cos(x)\) has a maximum at \(x = 0\) (\(\cos(0) = 1\)), \(y=-\cos(x)\) has a minimum at \(x = 0\) (\(-\cos(0)=-1\)), and \(y = -\sin(x)\) is a reflection of \(y=\sin(x)\) over the \(x\)-axis.

Step2: Analyze the given graph

Looking at the provided graph, at \(x = 0\), the function passes through the origin (since it crosses the \(y\)-axis at \(y = 0\)). Now, check the behavior around \(x = 0\). The standard \(y=\sin(x)\) has a positive slope at \(x = 0\) (increasing), but the given graph seems to have a negative slope at \(x = 0\) (decreasing). The function \(y=-\sin(x)\) has \(-\sin(0)=0\) and its derivative at \(x = 0\) is \(-\cos(0)=-1\) (negative slope), which matches the graph's behavior at \(x = 0\). Let's check other options:

• \(f(x)=\cos(x)\): At \(x = 0\), \(y = 1\), but the graph passes through \((0,0)\), so eliminate.
• \(f(x)=-\cos(x)\): At \(x = 0\), \(y=-1\), but the graph passes through \((0,0)\), so eliminate.
• \(f(x)=\sin(x)\): At \(x = 0\), slope is positive (increasing), but the graph is decreasing at \(x = 0\), so eliminate.
• \(f(x)=-\sin(x)\): At \(x = 0\), \(y = 0\), and the slope is negative (decreasing), which matches the graph.

Answer:

\(f(x)=-\sin(x)\) (the option with \(f(x)=-\sin(x)\))