Can someone solve 2(x-1)=10?
2 answers:
You might be interested in
Answer:
The function that could be the function described is;
Step-by-step explanation:
The given parameters of the cosine function are;
The period of the cosine function = 3
The maximum value of the cosine function = 20
The minimum value of the cosine function = 0
The general form of the cosine function is presented as follows;
y = A·cos(ω·x - ∅) + k
Where;
= The amplitude = Constant
The period, T = 2·π/ω
The phase shift, = ∅/ω
k = The vertical translation = Constant
Therefore, by comparison, we have;
T = 3 = 2·π/ω
∴ ω = 2·π/3
The range of value of the cosine of an angle are;
-1 ≤ cos(θ) ≤ 1
Therefore, when A = 10, cos(ω·x - ∅) = 1 (maximum value of cos(θ)) and k = 10, we have;
y = A × cos(ω·x - ∅) + k
y = 10 × 1 + 10 = 20 = The maximum value of the function
Similarly, when A = 10, cos(ω·x - ∅) = -1 (minimum value of cos(θ)) and k = 10, we get;
y = 10 × -1 + 10 = 0 = The minimum value of the function
Given that the function is a reflection of the parent function, we can have;
A = -10, cos(ω·x - ∅) = -1 (minimum value of cos(θ)) and k = 10, to get;
y = -10 × -1 + 10 = 20 = The maximum value of the function
Similarly, for cos(ω·x - ∅) = 1 we get;
y = -10 × 1 + 10 = 0 = The minimum value of the function
Therefore, the likely values of the function are therefore;
A = -10, k = 10
The function is therefore presented as follows;
y = -10 × cos(2·π/3·x) + 10