Answer:
Sure, based on the image you sent, the trigonometric equation that matches the graph is:
**y = ¾ sin(π(x + 1)) - 1**
Here's how we can analyze the graph to arrive at this equation:
**Sine function:** The graph exhibits up-and-down cycles, a characteristic of sine and cosine functions. Since the graph starts at its peak (y = ¾) and intersects the x-axis at x = 0, it most likely represents a sine function and not a cosine function.
**Amplitude:** The midline (horizontal line in the middle) of the graph is y = -1. The function oscillates between 1 unit above (y = ¾) and 2 units below (y = -2.75) the midline. So, the amplitude is the absolute value of half the difference between these extreme points, which is |(¾) - (-2.75)| = 3.05. However, note that the amplitude in a sine function is typically represented by a positive value, so we can rewrite this mathematically as:
a = ¾ (notice how this value appears directly in front of the sine term in the equation)
**Period:** One complete cycle of the graph spans across 4 units on the x-axis. This means the period is T = 4.
**Frequency:** The frequency (f) is the reciprocal of the period (T). Therefore, f = 1/T = 1/4. This means the function completes one cycle for every 4 units on the x-axis.
**Phase shift:** The graph doesn't start at its midline (y = -1) but rather at its peak (y = ¾). Then, it shifts 1 unit to the left before reaching its first zero crossing (x-axis intercept). A positive phase shift corresponds to a shift to the left. So, considering both these observations, there's a total phase shift of -1 units.
**Matching the equation to the graph:**
As discussed earlier, the general form for a sine function is:
y = a sin(bx + c) + d
where:
- a: amplitude (how far the function oscillates from the midline)
- b: frequency (how many cycles the function completes within a 2π unit interval) **Note that for this specific case, b = 2π/T**
- c: phase shift (horizontal shift to the right if positive, left if negative)
- d: midline (vertical shift)
In this case:
- a = ¾ (amplitude)
- b: Since the period is T = 4, b = 2π/T = 2π/4 = π/2 (frequency)
- c: -1 (phase shift to the left by 1 unit)
- d = -1 (midline)
**However, there's a mathematical property of sine functions that allows us to compress or stretch the graph horizontally.** Stretching the graph horizontally by a factor of k is equivalent to dividing the frequency (b) by k. In this case, the graph is stretched horizontally by a factor of 2 (compared to a standard sine function). Therefore, to compensate for this stretching and get the correct period, we need to divide the frequency (b) by 2.
Adjusted frequency (b') = b / 2 = (π/2) / 2 = π/4
**Incorporating these adjustments, the equation becomes:**
y = ¾ sin(π(x + (-1))) - 1 // Notice the phase shift (-1) is introduced within the sine term
**Simplifying the expression:**
y = ¾ sin(π(x + 1)) - 1
