Question

short answer
17 a. sketch at least one cycle of a sine curve with amplitude 2 and vertical shift of 3.

Explanation:

Step1: Recall the general form of a sine function

The general form of a sine function is \(y = A\sin(Bx - C)+D\), where \(A\) is the amplitude, \(D\) is the vertical shift. Given \(A = 2\) and \(D=3\), and for simplicity, we can take \(B = 1\) and \(C = 0\). So the function is \(y=2\sin(x)+3\).

Step2: Find key points

• Mid - line: \(y = 3\) (because of the vertical shift \(D = 3\)).
• Amplitude: The distance from the mid - line to the maximum and minimum values. The maximum value is \(y=3 + 2=5\) and the minimum value is \(y=3-2 = 1\).
• Period: For \(y=\sin(x)\), the period \(T = 2\pi\).
• Key points for \(y=\sin(x)\): \((0,0)\), \((\frac{\pi}{2},1)\), \((\pi,0)\), \((\frac{3\pi}{2},- 1)\), \((2\pi,0)\).
• Key points for \(y = 2\sin(x)+3\):
• When \(x = 0\), \(y=2\sin(0)+3=3\)
• When \(x=\frac{\pi}{2}\), \(y=2\sin(\frac{\pi}{2})+3=2\times1 + 3=5\)
• When \(x=\pi\), \(y=2\sin(\pi)+3=3\)
• When \(x=\frac{3\pi}{2}\), \(y=2\sin(\frac{3\pi}{2})+3=2\times(-1)+3 = 1\)
• When \(x = 2\pi\), \(y=2\sin(2\pi)+3=3\)

Step3: Sketch the graph

Plot the key points \((0,3)\), \((\frac{\pi}{2},5)\), \((\pi,3)\), \((\frac{3\pi}{2},1)\), \((2\pi,3)\) and connect them with a smooth curve that has the shape of a sine wave. The curve oscillates between \(y = 1\) and \(y = 5\) with a mid - line \(y = 3\) and a period of \(2\pi\).

Answer:

Sketch the sine curve \(y = 2\sin(x)+3\) using the key points \((0,3)\), \((\frac{\pi}{2},5)\), \((\pi,3)\), \((\frac{3\pi}{2},1)\), \((2\pi,3)\) and connecting them with a smooth wave - like curve that oscillates between \(y = 1\) and \(y = 5\) with a period of \(2\pi\).