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The height of the tide over time is shown on the graph. Which of the following equations can be used to reasonably model the data represented in the graph? Select two of the following that apply. There needs to be two selections!

The height of the tide over time is shown on the graph Which of the following equations can be used to reasonably model the data represented in the graph Select class=
The height of the tide over time is shown on the graph Which of the following equations can be used to reasonably model the data represented in the graph Select class=

Respuesta :

carlosego

Answer:

The correct options are the first and the fourth

[tex]f(x) = -3sin(\frac{\pi}{6}x) + 3[/tex]

and

[tex]f(x) = 3cos(\frac{\pi}{6}x+\frac{\pi}{2}) + 3[/tex]

Step-by-step explanation:

We can observe in the graph that when x = 0 then y = 3. Therefore f(0) must be equal to 3.

We also observe in the graph that f(x) = 0 when x = 3+12k, {3, 15, 27...}

where k is a Whole number k : {0, 1, 3, 4, 5....n}

The function that meets these two conditions is

[tex]f(x) = -3sin(\frac{\pi}{6}x) + 3[/tex]

and

[tex]f(x) = 3cos(\frac{\pi}{6}x+\frac{\pi}{2}) + 3[/tex]

We can check it by substituting x = 3+12k and verifying that f(x) = 0

[tex]f(3+12k) = -3sin(\frac{\pi}{6}(3+12k)) + 3[/tex]

[tex]f(3k) = -3sin(\frac{\pi}{2} + 2\pi(k)) + 3[/tex]

We know that [tex]sin(\frac{\pi}{2})) = 1[/tex] and [tex]sin(2k\pi) = 0[/tex]

Then

[tex]f(3+12k) = -3+ 3 = 0[/tex]

Also

[tex]f(3+12k) = 3cos(\frac{\pi}{6}(3k) + \frac{\pi}{2})+ 3[/tex]

We know that [tex]cos(\frac{\pi}{2}k) = -1[/tex]  and [tex]cos(2k\pi) = 1[/tex]

[tex]f(3+ 12k) = -3 +3 = 0[/tex]

Therefore The correct options are the first and the fourth

Answer:

Option 1 and 4 are the two functions.

Step-by-step explanation:

The given graph may be of sine function or cosine function.

So we start with sine function first.

f(x) = asin(bx+c) +d

Here the features of given graph are

1). Amplitude a = (6-0)/2 =3

2). Period = 12 therefore b = 2π/period = 2π/12 = π/6

3). Midline is y = 3

4). Horizontal shift c = 0

5). vertical shift d = 3 in upward direction

5). Since graph starts below the midline means the function will start with negative notation

So we design the function as y = -3sin(π/6+0)+3

Now for cosine function f(x) = a cos(bx+c)+d

1). Amplitude a = 3

2). Midline is y = 3

3). period = 2π therefore b = 2π/period = 2π/12 = π/6

4). Horizontal shift which is in right side, c = +(π/2)

5). Vertical shift d = 3

Therefore the cosine function will be f(x) = 3cos(π/6+π/2)+3

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