Math  /  Trigonometry

QuestionOne cycle of the graph of a trigonometric function of the form y=Asin(Bx)y=A \boldsymbol{\operatorname { s i n }}(B x) or y=Acos(Bx)y=A \boldsymbol{\operatorname { c o s }}(B x) is given. Determine the equation of the function represented by the following graph.
What function of the form y=Asin(Bx)y=A \sin (B x) or y=Acos(Bx),B>0y=A \cos (B x), B>0 is represented by the given graph? \square (bse integers or fractions for any numbers in the equation.)

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a trigonometric function (y=Asin(Bx)y = A \sin(Bx) or y=Acos(Bx)y = A \cos(Bx)) that matches the given graph. Watch out! Don't mix up sine and cosine, and make sure you get the *amplitude*, *period*, and *vertical shift* right!

STEP 2

1. Analyze the graph
2. Determine the amplitude *A*
3. Determine the vertical shift *D*
4. Determine the period and *B*
5. Choose sine or cosine and write the equation

STEP 3

Let's look at this graph!
We see it starts at a **maximum** value, which hints that it might be a cosine function.
It goes up to y=8y = 8 and down to y=2y = -2.
The graph completes one full cycle between x=0x = 0 and x=3π4x = \frac{3\pi}{4}.

STEP 4

The **amplitude** is the distance from the **midline** to the **maximum** (or **minimum**) value.
The midline is halfway between the highest and lowest points.
We calculate this by finding the average of the maximum and minimum values: 8+(2)2=62=3\frac{8 + (-2)}{2} = \frac{6}{2} = 3.
The distance from the midline (y=3y=3) to the maximum (y=8y=8) is 83=58 - 3 = 5.
So, our **amplitude** AA is **5**.

STEP 5

The **vertical shift** is the value of the midline.
We already found that the midline is at y=3y = 3, so our vertical shift DD is **3**.

STEP 6

The **period** is the length of one full cycle of the graph.
We can see that the graph completes one cycle from x=0x = 0 to x=3π4x = \frac{3\pi}{4}.
So, the period is 3π4\frac{3\pi}{4}.

STEP 7

Now, we know the period is related to BB by the formula Period=2πBPeriod = \frac{2\pi}{B}.
We have 3π4=2πB\frac{3\pi}{4} = \frac{2\pi}{B}.
To solve for BB, we can cross-multiply: 3πB=2π43\pi \cdot B = 2\pi \cdot 4, which simplifies to 3πB=8π3\pi B = 8\pi.
Now, we divide both sides by 3π3\pi to isolate BB: B=8π3π=83B = \frac{8\pi}{3\pi} = \frac{8}{3}.
So, B=83B = \frac{8}{3}.

STEP 8

Since the graph starts at a **maximum**, we'll use cosine.
A standard cosine graph starts at its maximum, while sine starts at the midline.

STEP 9

We found A=5A = 5, B=83B = \frac{8}{3}, and D=3D = 3.
Since we are using cosine, the general form is y=Acos(Bx)+Dy = A \cos(Bx) + D.
Plugging in our values, we get y=5cos(83x)+3y = 5 \cos\left(\frac{8}{3}x\right) + 3.

STEP 10

The equation of the function is y=5cos(83x)+3y = 5 \cos\left(\frac{8}{3}x\right) + 3.

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