Question

standard 15 homework
score: 11.67/21 answered: 13/20
question 14
sketch a graph of the function $f(x)=-2\sin(\frac{\pi}{2}x)-3$
question help: video message instructor post to forum
submit question

Explanation:

Step1: Determine the amplitude

The general form of a sine function is \(y = A\sin(Bx - C)+D\). For the function \(f(x)=- 2\sin(\frac{\pi}{2}x)-3\), the amplitude \(|A| = |-2|=2\).

Step2: Determine the period

The period of a sine function \(y=\sin(Bx)\) is \(T=\frac{2\pi}{|B|}\). Here \(B = \frac{\pi}{2}\), so \(T=\frac{2\pi}{\frac{\pi}{2}}=4\).

Step3: Determine the vertical shift

The vertical shift is \(D=-3\). This means the graph of \(y = \sin(x)\) is shifted down by 3 units.

Step4: Determine the reflection

Since \(A=-2\), the graph is reflected about the \(x\) - axis.

To sketch the graph:

• Start with the basic sine function \(y = \sin(x)\) which has key points \((0,0)\), \((\frac{\pi}{2},1)\), \((\pi,0)\), \((\frac{3\pi}{2},-1)\), \((2\pi,0)\)
• For \(y=-2\sin(\frac{\pi}{2}x)-3\):
• When \(x = 0\), \(y=-2\sin(0)-3=- 3\)
• When \(x = 1\), \(y=-2\sin(\frac{\pi}{2})-3=-2 - 3=-5\)
• When \(x = 2\), \(y=-2\sin(\pi)-3=-3\)
• When \(x = 3\), \(y=-2\sin(\frac{3\pi}{2})-3=2 - 3=-1\)
• When \(x = 4\), \(y=-2\sin(2\pi)-3=-3\)

Plot these key points \((0,-3)\), \((1,-5)\), \((2,-3)\), \((3,-1)\), \((4,-3)\) and then connect them with a smooth sine - wave curve. The graph oscillates between \(y=-1\) (maximum value: \(-2\times(- 1)-3=-1\)) and \(y=-5\) (minimum value: \(-2\times1 - 3=-5\)) with a period of \(4\) and is shifted down by \(3\) units.

Answer:

Sketch the graph using the key points \((0,-3)\), \((1,-5)\), \((2,-3)\), \((3,-1)\), \((4,-3)\) and connecting them with a smooth sine - wave curve that oscillates between \(y = - 1\) and \(y=-5\) with a period of \(4\) and is shifted down by \(3\) units.