Question

write the equation of the trigonometric function shown in the graph.

Explanation:

Step1: Identify the type of trigonometric function

The graph appears to be a sine function (since it passes through the origin-like point and has the shape of a sine wave). The general form of a sine function is \( y = A\sin(Bx + C) + D \).

Step2: Determine the amplitude (A)

The amplitude is the maximum distance from the midline to the peak (or trough). Looking at the graph, the maximum value is 2 and the minimum is -4? Wait, no, wait. Wait, the midline: let's check the vertical shift. Wait, the graph oscillates between, let's see, the peaks are at y=2 and troughs at y=-4? Wait, no, maybe I misread. Wait, looking at the graph, the peak is at y=2, and the trough is at y=-4? Wait, no, the distance from midline to peak: let's find the midline. The midline (D) is the average of the maximum and minimum. Let's see, the maximum y-value is 2, the minimum is -4? Wait, no, maybe the graph is a sine function with amplitude. Wait, actually, looking at the graph, from the trough to the peak: the trough is at y=-4, peak at y=2? No, that can't be. Wait, maybe I made a mistake. Wait, the graph: when x=0, y is between -1 and 0? Wait, no, the graph crosses the y-axis at y=-1? Wait, maybe the amplitude is 3? Wait, let's re-examine. The peak is at y=2, trough at y=-4. The midline (D) is (2 + (-4))/2 = (-2)/2 = -1. So the midline is y = -1. Then the amplitude (A) is the distance from midline to peak: 2 - (-1) = 3. So A = 3.

Step3: Determine the period (and thus B)

The period of a sine function is \( \frac{2\pi}{|B|} \). Let's find the period. Looking at the graph, from one peak to the next peak: from \( -\frac{3\pi}{4} \) to \( \frac{\pi}{4} \), the distance is \( \frac{\pi}{4} - (-\frac{3\pi}{4}) = \pi \). So the period is \( \pi \). Then \( \frac{2\pi}{|B|} = \pi \), so \( |B| = 2 \), so B = 2 (since the function is increasing at x=0? Wait, at x=0, the function is increasing (since it goes from trough to peak after x=0? Wait, at x=0, the graph is below the midline and increasing? Wait, the midline is y=-1. At x=0, y is -1? Wait, no, the graph crosses the y-axis at y=-1? Wait, maybe the phase shift (C) is 0. Wait, let's check the phase shift. The general form: \( y = A\sin(Bx) + D \). Wait, if B=2, D=-1, A=3. Let's test. So \( y = 3\sin(2x) - 1 \). Let's check at x = \( \frac{\pi}{4} \): \( 3\sin(2 \times \frac{\pi}{4}) - 1 = 3\sin(\frac{\pi}{2}) - 1 = 3(1) - 1 = 2 \), which matches the peak at x = \( \frac{\pi}{4} \), y=2. At x = \( \frac{3\pi}{4} \), trough: \( 3\sin(2 \times \frac{3\pi}{4}) - 1 = 3\sin(\frac{3\pi}{2}) - 1 = 3(-1) - 1 = -4 \), which matches the trough. At x = 0: \( 3\sin(0) - 1 = -1 \), which matches the y-intercept. So that works.

Step4: Determine the phase shift (C) and vertical shift (D)

We already found D = -1 (midline at y=-1). The phase shift: since the function is \( y = 3\sin(2x) - 1 \), there is no phase shift (C=0) because the sine function starts at the midline (y=-1) when x=0, and increases, which matches the graph (at x=0, y=-1, and then increases towards the peak at x=π/4).

So putting it all together: \( y = 3\sin(2x) - 1 \)? Wait, but let's check again. Wait, when x=π/4, 2x=π/2, sin(π/2)=1, so 3(1) -1=2, correct. When x=3π/4, 2x=3π/2, sin(3π/2)=-1, so 3(-1)-1=-4, correct. The period is π, which matches. The amplitude is 3, midline y=-1. So the equation is \( y = 3\sin(2x) - 1 \)? Wait, but let's check the graph again. Wait, the graph at x=0 is at y=-1, then goes up to x=π/4 (peak at y=2), then down to x=3π/4 (trough at y=-4), then up again. That matches. Alternatively, maybe it's a sine function with amplitude 3, period π, vertical shift -1, and no phase shift. So the equation is \( y = 3\sin(2x) - 1 \). Wait, but let's confirm the period. From -3π/4 to π/4 is π, so period is π, so B=2. Amplitude: 3, vertical shift -1. So yes, that seems correct.

Wait, but another way: maybe the function is \( y = 3\sin(2x) - 1 \). Let's check x=π/2: 2x=π, sin(π)=0, so y= -1. Which is on the midline. Correct. x=π: 2x=2π, sin(2π)=0, y=-1. Correct. So the equation is \( y = 3\sin(2x) - 1 \).

Answer:

\( y = 3\sin(2x) - 1 \)