f(x)=6\\cos(x)\nf(x)=3\\cos(x)+3\nf(x)=6\\sin(x)\nf(x)=3\\sin(x)+3
f(x)=6\\cos(x)\nf(x)=3\\cos(x)+3\nf(x)=6\\sin(x)\nf(x)=3\\sin(x)+3
Answer
Answer:
B. $f(x)=3\cos(x)+3$
Explanation:
Step1: Analyze the amplitude
The general form of a cosine - sine function is $y = A\sin(Bx - C)+D$ or $y = A\cos(Bx - C)+D$. The amplitude $|A|$ is half of the vertical distance between the maximum and minimum values of the function. The maximum value of the given function is 6 and the minimum is 0. So, the amplitude $A=\frac{6 - 0}{2}=3$.
Step2: Analyze the vertical shift
The vertical shift $D$ is the mid - value between the maximum and minimum values. $D=\frac{6 + 0}{2}=3$.
Step3: Determine the function type
The function has a value of 6 at $x = 0$. For a cosine function $y=A\cos(x)+D$, when $x = 0$, $y=A + D$. For a sine function $y=A\sin(x)+D$, when $x = 0$, $y = D$. Since when $x = 0$, $y=6$ and $A = 3,D = 3$, substituting into $y=A\cos(x)+D$ gives $y=3\cos(x)+3$ (because when $x = 0$, $\cos(0)=1$ and $y=3\times1 + 3=6$), while for $y = A\sin(x)+D$, when $x = 0$, $y=D = 3\neq6$. So the function is $f(x)=3\cos(x)+3$.