Question

write the equation for the trigonometric function based on its graph. (1 point) \\(\bigcirc\\) \\(y = \sin(x) - 1\\) \\(\bigcirc\\) \\(y = \cos(x) - 1\\) \\(\bigcirc\\) \\(y = \cos x\\) \\(\bigcirc\\) \\(y = \cos(x) - 2\\)

Explanation:

Step1: Analyze the base function

The graph has a maximum at \( x = 0 \), which is characteristic of the cosine function \( y=\cos(x) \) (since \( \cos(0) = 1 \)), while \( \sin(0)=0 \), so we can eliminate the \( y = \sin(x)-1 \) option.

Step2: Analyze the vertical shift

The standard \( y = \cos(x) \) has a range of \([-1, 1]\). The given graph has a minimum value of \(-2\) (since the troughs are at \( y=-2 \)). The vertical shift \( k \) in the form \( y=\cos(x)+k \) can be found by looking at the midline. The midline of \( y = \cos(x) \) is \( y = 0 \), but here the midline is \( y=-1 \)? Wait, no, let's check the minimum. The standard \( \cos(x) \) minimum is \(-1\), here the minimum is \(-2\), so the vertical shift is \( -2 \)? Wait, no, let's check the graph again. Wait, at \( x = 0 \), the graph has a value of \( 0 \)? Wait, no, the graph at \( x = 0 \) is at \( y = 0 \)? Wait, no, the standard \( \cos(x) \) at \( x = 0 \) is \( 1 \). Wait, maybe I made a mistake. Wait, the graph: let's check the key points. The standard \( y=\cos(x) \) has a maximum at \( (0, 1) \), minimum at \( (\pi, -1) \). The given graph: at \( x = 0 \), the value is \( 0 \); at \( x=\pi \), the value is \( -2 \); at \( x = 2\pi \), the value is \( 0 \). Wait, let's check the options. The options are \( y=\sin(x)-1 \), \( y=\cos(x)-1 \), \( y=\cos(x) \), \( y=\cos(x)-2 \). Let's test \( y=\cos(x)-1 \): at \( x = 0 \), \( \cos(0)-1=1 - 1=0 \), which matches the graph's \( x = 0 \) point. At \( x=\pi \), \( \cos(\pi)-1=-1 - 1=-2 \), which matches the minimum. At \( x = 2\pi \), \( \cos(2\pi)-1=1 - 1=0 \), which matches. Wait, but earlier I thought the minimum was \(-2\), and \( \cos(x)-1 \) has a minimum of \(-2\) (since \( \cos(x) \) minimum is \(-1\), so \(-1 - 1=-2\)). Wait, but let's check the other options. \( y=\cos(x)-2 \) would have a minimum of \(-3\) (since \( \cos(x) \) minimum is \(-1\), so \(-1 - 2=-3\)), but the graph's minimum is \(-2\), so that's out. \( y=\cos(x) \) has a minimum of \(-1\), but the graph's minimum is \(-2\), so out. \( y=\sin(x)-1 \): at \( x = 0 \), \( \sin(0)-1=0 - 1=-1 \), which does not match the graph's \( x = 0 \) value (which is \( 0 \)). So the correct option is \( y=\cos(x)-1 \)? Wait, no, wait the graph at \( x = 0 \): the red curve at \( x = 0 \) is at \( y = 0 \). \( \cos(0)-1=1 - 1=0 \), correct. At \( x=\pi \), \( \cos(\pi)-1=-1 - 1=-2 \), correct. At \( x = 2\pi \), \( \cos(2\pi)-1=1 - 1=0 \), correct. So the equation is \( y=\cos(x)-1 \). Wait, but let's check the options again. The options are:

• \( y = \sin(x) - 1 \)
• \( y = \cos(x) - 1 \)
• \( y = \cos x \)
• \( y = \cos(x) - 2 \)

Testing \( y=\cos(x)-1 \):

• At \( x = 0 \): \( \cos(0)-1 = 1 - 1 = 0 \) (matches the graph's \( x=0 \) point)
• At \( x = \pi \): \( \cos(\pi)-1 = -1 - 1 = -2 \) (matches the minimum at \( x=\pi \))
• At \( x = 2\pi \): \( \cos(2\pi)-1 = 1 - 1 = 0 \) (matches the graph's \( x=2\pi \) point)

Other options:

• \( y=\sin(x)-1 \): At \( x=0 \), \( \sin(0)-1 = -1

eq 0 \)

• \( y=\cos(x) \): At \( x=0 \), \( \cos(0)=1

eq 0 \); at \( x=\pi \), \( \cos(\pi)=-1
eq -2 \)

• \( y=\cos(x)-2 \): At \( x=0 \), \( \cos(0)-2 = -1

eq 0 \); at \( x=\pi \), \( \cos(\pi)-2 = -3
eq -2 \)

So the correct option is \( y = \cos(x) - 1 \). Wait, but wait the graph's minimum is at \( y=-2 \), and \( \cos(x)-1 \) has a minimum of \(-2\) (since \( \cos(x) \) ranges from \(-1\) to \( 1 \), so \( \cos(x)-1 \) ranges from \(-2\) to \( 0 \)), which matches the graph (since the graph goes from \( 0 \) (max) to \(-2\) (min)). Yes, that's cor…

Answer:

\( y = \cos(x) - 1 \) (the option corresponding to this equation, which is the second option: \( y = \cos(x) - 1 \))