To create an equation in the form y = a sin [k (x - d)] + c, where the solutions are at x = 1 and x = 2, we need to choose appropriate values for a, k, d, and c.
Considering that the solutions are at x = 1 and x = 2, we can deduce that the midline of the graph (the horizontal line y = c) will be the average of these two values, which is 1.5. Therefore, we can set c = 1.5.
Now let's focus on the amplitude of the graph. Since the sine function has an amplitude of 1, we can choose a value for a to adjust the amplitude accordingly. Let's set a = 0.5 to make the amplitude of the graph smaller.
Next, we need to determine the period of the graph, which corresponds to the value of k in the equation. Since we want the solutions to occur at x = 1 and x = 2, the period of the graph needs to be 1 unit. Thus, we set k = 2π.
Finally, we need to decide on the phase shift of the graph, represented by d in the equation. The phase shift indicates the displacement of the graph along the x-axis. Since we want the first solution at x = 1, we can set d = 1.
Putting all this together, the equation can be written as:
y = 0.5 sin[2π (x - 1)] + 1.5.
This equation satisfies the given conditions and has solutions only at x = 1 and x = 2.
Provide a trigonometric equation. Considering only the space between 𝑥 = 0 and 2𝜋, the equation must only have solutions at 𝑥 = 1 and 𝑥 = 2. Explain your thought process and the work you did to create the equation. You may round decimal values to 3 places.
I was thinking y=cos(x / 50)(x-1)(x-2), however I would like, if possible, some help to create one in a y= a sin,cos [k (x-d) ] + c format. Thanks.
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