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Math

Math Snap

PROBLEM

One 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.)

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.

SOLUTION

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

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