Question

consider the function ( y = 2sin(x) ) for ( 0^{circ}leq xleq 360^{circ} ).

1. graph the function:

plot the function ( y = 2sin(x) ) on a coordinate plane. label the x - axis as \angle (degrees)\ and the y - axis as \y - value\. mark the coordinates of all key points where the graph intersects the x - axis, reaches its maximum, or minimum within the given interval.

Explanation:

Step1: Find x - intercepts

Set \(y = 0\), solve \(2\sin(x)=0\), \(\sin(x)=0\). Using the unit - circle or sine function properties, \(x = k\cdot180^{\circ},k\in\mathbb{Z}\). In the interval \(0^{\circ}\leq x\leq360^{\circ}\), \(x = 0^{\circ},180^{\circ},360^{\circ}\)

Step2: Find maximum value

The general form of \(y = A\sin(x)+B\) (here \(B = 0\), \(A=2\)). The maximum of \(\sin(x)\) is \(1\). Substitute \(x = 90^{\circ}\) (where \(\sin(x)\) is maximum) into \(y = 2\sin(x)\), \(y=2\)

Step3: Find minimum value

The minimum of \(\sin(x)\) is \(-1\). Substitute \(x = 270^{\circ}\) (where \(\sin(x)\) is minimum) into \(y = 2\sin(x)\), \(y=-2\)

To graph:

1. Draw the coordinate plane with \(x\) - axis labeled "Angle (degrees)" and \(y\) - axis labeled "y - value"
2. Plot the points \((0^{\circ},0)\), \((90^{\circ},2)\), \((180^{\circ},0)\), \((270^{\circ},-2)\), \((360^{\circ},0)\)
3. Connect the points with a smooth curve that has the shape of a sine wave.

Answer:

To graph \(y = 2\sin(x)\) for \(0^{\circ}\leq x\leq360^{\circ}\):

• Intercepts:
• When \(y = 0\), \(2\sin(x)=0\), so \(\sin(x) = 0\). Then \(x = 0^{\circ},180^{\circ},360^{\circ}\). The points are \((0^{\circ},0)\), \((180^{\circ},0)\), \((360^{\circ},0)\)
• Maximum:
• The maximum value of \(\sin(x)\) is \(1\). When \(\sin(x)=1\), \(x = 90^{\circ}\). Then \(y=2\times1 = 2\). The point is \((90^{\circ},2)\)
• Minimum:
• The minimum value of \(\sin(x)\) is \(- 1\). When \(\sin(x)=-1\), \(x = 270^{\circ}\). Then \(y=2\times(-1)=-2\). The point is \((270^{\circ},-2)\)