Question

standard 15 homework
score: 10.67/21 answered: 12/20
question 13
sketch a graph of the function ( f(x)=3 cos (x)-3 )
clear all draw:
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Explanation:

Step1: Identify the standard cosine function

The standard cosine function is \(y = \cos(x)\), which has an amplitude of \(1\), a period of \(2\pi\), a maximum value of \(1\) at \(x = 2k\pi,k\in\mathbb{Z}\), and a minimum value of \(- 1\) at \(x=(2k + 1)\pi,k\in\mathbb{Z}\), and it passes through the point \((0,1)\).

Step2: Analyze the transformation for \(y = 3\cos(x)-3\)

• Amplitude transformation: For the function \(y = A\cos(x)+B\), the amplitude is \(|A|\). Here \(A = 3\), so the amplitude of \(y=3\cos(x)-3\) is \(3\). The maximum value of \(y = \cos(x)\) is \(1\), and for \(y = 3\cos(x)\) it is \(3\times1=3\), and for \(y = 3\cos(x)-3\) it is \(3 - 3=0\). The minimum value of \(y=\cos(x)\) is \(-1\), for \(y = 3\cos(x)\) it is \(3\times(- 1)=-3\), and for \(y = 3\cos(x)-3\) it is \(-3-3=-6\).
• Vertical - shift transformation: For the function \(y = A\cos(x)+B\), the vertical shift is \(B\). Here \(B=-3\).
• Period: The period of \(y = A\cos(x)+B\) is the same as the period of \(y=\cos(x)\) since there is no horizontal - scaling factor (\(y = A\cos(Cx)+B\), and here \(C = 1\)). The period \(T=\frac{2\pi}{|C|}=2\pi\)

Step3: Find key points

• When \(x = 0\): \(y=3\cos(0)-3=3\times1 - 3=0\)
• When \(x=\frac{\pi}{2}\): \(y = 3\cos(\frac{\pi}{2})-3=3\times0-3=-3\)
• When \(x=\pi\): \(y=3\cos(\pi)-3=3\times(-1)-3=-6\)
• When \(x=\frac{3\pi}{2}\): \(y = 3\cos(\frac{3\pi}{2})-3=3\times0-3=-3\)
• When \(x = 2\pi\): \(y=3\cos(2\pi)-3=3\times1-3=0\)

Plot these key points \((0,0),(\frac{\pi}{2},-3),(\pi,-6),(\frac{3\pi}{2},-3),(2\pi,0)\) and then connect them with a smooth curve that has a period of \(2\pi\) to sketch the graph of \(y = 3\cos(x)-3\).

Answer:

Sketch the graph using the key - points \((0,0),(\frac{\pi}{2},-3),(\pi,-6),(\frac{3\pi}{2},-3),(2\pi,0)\) and the fact that the function has a period of \(2\pi\) and is a cosine - type curve.