Math / Trigonometry

QuestionSuppose $y=3 \sin (4(t+13))-6$ . In your answers, enter pi for $\pi$ . (a) The midline of the graph is the line with equation $y=-6$ help (equations) (b) The amplitude of the graph is 3 help (numbers) (c) The period of the graph is $\pi$ help (numbers)
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

1. The function given is $y = 3 \sin(4(t + 13)) - 6$ .
2. The function is a transformation of the basic sine function.
3. The general form of a sine function is $y = A \sin(B(t - C)) + D$ , where: - $A$ is the amplitude. - $B$ affects the period. - $C$ is the horizontal shift. - $D$ is the vertical shift (midline).

STEP 2

1. Identify the midline of the graph.
2. Determine the amplitude of the graph.
3. Calculate the period of the graph.

STEP 3

The midline of a sine function is determined by the vertical shift $D$ .
For the function $y = 3 \sin(4(t + 13)) - 6$ , the midline is given by the constant term outside the sine function, which is $-6$ .
Thus, the equation of the midline is:
$y = -6$

STEP 4

The amplitude of a sine function is the absolute value of the coefficient $A$ in front of the sine term.
For the function $y = 3 \sin(4(t + 13)) - 6$ , the amplitude is the absolute value of 3, which is:
$\text{Amplitude} = 3$

STEP 5

The period of a sine function is calculated using the coefficient $B$ inside the sine function. The formula for the period is:
$\text{Period} = \frac{2\pi}{B}$
For the function $y = 3 \sin(4(t + 13)) - 6$ , $B = 4$ .
$\text{Period} = \frac{2\pi}{4} = \frac{\pi}{2}$
The midline of the graph is $y = -6$ , the amplitude is $3$ , and the period is $\frac{\pi}{2}$ .

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