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nadya68 [22]
3 years ago
13

The graph of y=cos⁡x is transformed to y=a cos ⁡(x−c)+d by a vertical compression by a factor of 1/3 and a translation 2 units d

own. The new equation is:
y=3cos⁡x+2


y=1/3cos⁡(x−2)


y=1/3cos⁡x−2


y=1/3cos⁡x+2

Mathematics
2 answers:
rodikova [14]3 years ago
5 0

ANSWER

y =  \frac{1}{3}  \cos(x)  - 2

EXPLANATION

If the graph of y=cos⁡x is transformed to y=a cos ⁡(x−c)+d by a vertical compression by a factor of 1/3 and a translation 2 units down,

then

a=1/3

and d=-2.

The 'c' is a phase shift since it is not given, it means it is zero.

Therefore the new equation is:

y=1/3cos⁡(x-0)−2

This simplifies to:

y=1/3cos⁡x−2

The correct option is C.

SIZIF [17.4K]3 years ago
4 0

Answer:

The new equation is y = 1/3 cos(x) - 2 ⇒ 3rd answer

Step-by-step explanation:

* Lets revise the trigonometry transformation

- If the equation is y = a cos(x - c) + d

# a is the scale factor of a vertical stretch or compression  

# c is the phase shift  (negative is to the right, positive is to the left)

# d is the vertical shift

- If y = cos(x)

∴ a = 1 , c = 0 , d = 0

* Now lets solve the problem

∵ There is a vertical compression by a factor of 1/3

∴ a = 1/3

∵ There is a translation 2 units down (vertical translation)

∴ d = -2

∵ There is now phase shift (horizontal translation)

∴ c = 0

* Now lets write the new equation

∴ y = 1/3 cos(x) - 2

* For more understand look to the attached color graph

- The red is y = cos(x)

- The blue is y = 1/3 cos(x) - 2

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Explanation:

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<u>Example Problem</u>

For the example attached, we are finding the area under the sine curve on the interval [1, 4] using 6 subintervals. We are using a rectangle whose height matches the function at the left side of the rectangle. We say this is a <em>left sum</em>.

When rectangles are used, other choices often seen are <em>right sum</em>, or <em>midpoint sum</em> (where the midpoint of the rectangle matches the function value at that point).

Each term of the sum is the area of the rectangle. That is the product of the rectangle's height and its width. We have chosen the width of the rectangle (the "subinterval") to be 1/6 of the width of the interval [1, 4], so each rectangle is (4-1)/6 = 1/2 unit wide.

The height of each rectangle is the function value at its left edge. In the example, we have defined the function x₁(j) to give us the x-value at the left edge of subinterval j. Then the height of the rectangle is f(x₁(j)).

We have factored the rectangle width out of the sum, so our sum is simply the heights of the left edges of the 6 subintervals. Multiplying that sum by the subinterval width gives our left sum r₁. (It is not a very good approximation of the integral.)

The second and third attachments show a <em>right sum</em> (r₂) and a <em>midpoint sum</em> (r₃). The latter is the best of these approximations.

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<u>Other Rules</u>

Described above and shown in the graphics are the use of <em>rectangles</em> for elements of the summation. Another choice is the use of <em>trapezoids</em>. For this, the corners of the trapezoid match the function value on both the left and right edges of the subinterval.

Suppose the n subinterval boundaries are at x0, x1, x2, ..., xn, so that the function values at those boundaries are f(x0), f(x1), f(x2), ..., f(xn). Using trapezoids, the area of the first trapezoid would be ...

  a1 = (f(x0) +f(x1))/2·∆x . . . . where ∆x is the subinterval width

  a2 = (f(x1) +f(x2))/2·∆x

We can see that in computing these two terms, we have evaluated f(x1) twice. We also see that f(x1)/2 contributes twice to the overall sum.

If we collapse the sum a1+a2+...+an, we find it is ...

  ∆x·(f(x0)/2 + f(x1) +f(x2) + ... +f(x_n-1) + f(xn)/2)

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<em>Comment on mechanics</em>

As you can tell from the attachments, it is convenient to let a graphing calculator or spreadsheet compute the sum. If you need to see the interval boundaries and the function values, a spreadsheet may be preferred.

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