Question

The graph of a trigonometric function oscillates between y=1 and y=-7. It reaches its
maximum at x=pi and its minimum at x=3pi. Which of the following could be the equation of the function?

Answer 1

Answer:

y = 4 sin(½ x) − 3

Step-by-step explanation:

The function is either sine or cosine:

y = A sin(2π/T x) + C

y = A cos(2π/T x) + C

where A is the amplitude, T is the period, and C is the midline.

The midline is the average of the min and max:

C = (1 + -7) / 2

C = -3

The amplitude is half the difference between the min and max:

A = (1 − -7) / 2

A = 4

The maximum is at x = π, and the minimum is at x = 3π.  The difference, 2π, is half the period.  So T = 4π.

Plugging in, the options are:

y = 4 sin(½ x) − 3

y = 4 cos(½ x) − 3

Since the maximum is at x = π, this must be a sine wave.

y = 4 sin(½ x) − 3

Answer 2

The trigonometric function is of Sine, at it can be given as $y = 4\ Sin(\dfrac{1}{2} x)-3$.

What is the trigonometric function of Sine and Cosine?

As it is given to us that the given graph is of a trigonometric function, therefore, the function is either sine or cosine, and we know that the general equation of sine and cosine is given as,

$\rm y = A\ Sin(\dfrac{2\pi}{T} x) + C\\\\y = A\ Cos(\dfrac{2\pi}{T} x) + C$

where A is the amplitude, T is the period, and C is the midline of the graph.

As we know that the amplitude is half the difference between the min and max oscillation of the graph, also, the function is minimum oscillation is at -7, while the maximum is at 1. therefore, the amplitude of the graph can be written as,

$A = \dfrac{a-b}{2}\\\\A = \dfrac{1 - (-7)}{2}\\\\A = 4$

Also, the midline is the average of the min and max oscillation of the graph, also, the function is minimum oscillation is at -7, while the maximum is at 1. therefore, the midline of the graph can be written as,

$C = \dfrac{a+b}{2}\\\\C = \dfrac{1 + (-7)}{2}\\\\C = -3$

We know that the period of oscillation is maximum is at x = π, and the minimum is at x = 3π.  The difference is 2π which is half the period. Therefore,

T = 4π.

Now as we know that the function is maximum at π, as we know that the value of Sine is maximum at π while the value of Cos is minimum at π. therefore, substituting the values in the general equation of Sine,

$\rm y = 4\ Sin(\dfrac{2\pi}{4 \pi} x)-3\\\\\rm y = 4\ Sin(\dfrac{1}{2} x)-3$

Hence, the trigonometric function is of Sine, at it can be given as $y = 4\ Sin(\dfrac{1}{2} x)-3$.

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