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4.1 homework question 3 fith grade
2 answers:
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Question # 13
Answer:
The required equation for the given function is <em>y = 4sin(x/2+2π/3) -2 , as shown attached graph diagram.</em>
<em>Step-by-step explanation: </em>
As the general sine function is given by
.......[A]
- amplitude = a
- period = 2π ÷ b
- Phase shift = -c ÷ b
- Vertical shift = d
As in the question,
- amplitude = a = 4
- period = 4π
- phase shift = -4π/3
- Vertical shift = d = -2
As
period = 2π ÷ b
b = 2π/period
b = 2π/4π ∵ period = 4π
b = 1/2
Also
Phase shift = -c/b
-4π/3 = -c/b ∵ phase shift = -4π/3
4π/3 = c/b
c = b × 4π/3
c = 1/2 × 4π/3
c = 4π/6
c = 2π/3
So, putting Amplitude ⇒ a = 4, Vertical shift ⇒ d = -2, b = 1/2 ,
and c = 2π/3 in Equation [A] would bring us the required equation for the given function.
y = 4sin(x/2+2π/3)+(-2)
y = 4sin(x/2+2π/3) -2
<em>Note: The graph is also shown in attached diagram.</em>
Question # 14
<em>Answer:</em>
The required equation for the given function is y = cot(x+π/3)+2, as shown in attached graph diagram.
<em>Step-by-step explanation: </em>
As the general cotangent function is given by
.......[A]
- amplitude = a
- period = π ÷ b
- Phase shift = -c ÷ b
- Vertical shift = d
As in the question,
- period = π
- phase shift = -π/3
- Vertical shift = d = 2
As
period = π ÷ b
b = π/period
b = π/π ∵ period = 4π
b = 1
Also
Phase shift = -c/b
-π/3 = -c/b ∵ phase shift = -π/3
π/3 = c/b
c = b × π/3
c = 1 × π/3
c = π/3
So, putting vertical shift ⇒ d = 2, b = 1 and c = π/3 in Equation [A] would bring us the required equation for the given function.
y = cot(x+π/3)+2
<em>Note: The graph is also shown in attached diagram.</em>
Keywords: amplitude, period , phase shift , vertical shift
Learn more about trigonometric functions of equations from brainly.com/question/2643311
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