Question

which function is graphed below?
(graph of a sinusoidal function with x-axis from 0 to 2π, y-axis from -2 to 2, passing through (0,0), peaking at (π/2,1), crossing x-axis at (π,0), trough at (3π/2,-1), and crossing x-axis at (2π,0))
options:

• f(x) = -cos(x)
• f(x) = cos(x)
• f(x) = sin(x)
• f(x) = -sin(x)

Explanation:

Step1: Recall basic trigonometric graphs

The standard sine function \( y = \sin(x) \) has a period of \( 2\pi \), passes through the origin \((0,0)\), reaches a maximum of \( 1 \) at \( \frac{\pi}{2} \), a minimum of \( -1 \) at \( \frac{3\pi}{2} \), and crosses the x - axis at \( 0,\pi,2\pi,\cdots \). The standard cosine function \( y=\cos(x) \) has a period of \( 2\pi \), has a maximum of \( 1 \) at \( x = 0 \), a minimum of \( - 1 \) at \( x=\pi \), and crosses the x - axis at \( \frac{\pi}{2},\frac{3\pi}{2},\cdots \).

Step2: Analyze the given graph

• The given graph passes through the origin \((0,0)\). The function \( y = \cos(x) \) has a value of \( 1 \) at \( x = 0 \), so we can eliminate \( f(x)=\cos(x) \) and \( f(x)=-\cos(x) \) (since \( y =-\cos(x) \) has a value of \( - 1 \) at \( x = 0 \)).
• Now we have two options left: \( f(x)=\sin(x) \) and \( f(x)=-\sin(x) \). Let's check the value of the function at \( x=\frac{\pi}{2} \). For \( y = \sin(x) \), when \( x=\frac{\pi}{2} \), \( y = 1 \). For \( y=-\sin(x) \), when \( x = \frac{\pi}{2} \), \( y=-1 \). Looking at the graph, at \( x=\frac{\pi}{2} \) (the peak of the curve), the y - value is \( 1 \). So the function is \( f(x)=\sin(x) \).

Answer:

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