Question

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

1. graph the function:

plot the function ( y = 2 sin ( 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: Analyze the general form of the sine function

The general form of a sine function is \(y = A\sin(Bx - C)+D\). For the function \(y = 2\sin(x)\), we have \(A = 2\), \(B = 1\), \(C = 0\), and \(D = 0\). The amplitude \(|A|=2\), the period \(T=\frac{360^{\circ}}{|B|}=360^{\circ}\), the phase - shift \(=\frac{C}{B}=0^{\circ}\), and the vertical - shift \(D = 0\).

Step2: Find the x - intercepts

Set \(y = 0\), so \(2\sin(x)=0\). Then \(\sin(x)=0\). Using the unit - circle or the sine - function properties, \(x = 0^{\circ}+n\cdot180^{\circ}\), \(n\in\mathbb{Z}\). In the interval \(0^{\circ}\leq x\leq360^{\circ}\), \(x = 0^{\circ},180^{\circ},360^{\circ}\). The coordinates of the x - intercepts are \((0^{\circ},0)\), \((180^{\circ},0)\), \((360^{\circ},0)\).

Step3: Find the maximum and minimum points

The maximum value of \(y = A\sin(x)\) occurs when \(\sin(x)=1\). So \(y = 2\times1 = 2\), and \(x = 90^{\circ}+n\cdot360^{\circ}\). In the given interval \(x = 90^{\circ}\), and the coordinate is \((90^{\circ},2)\).
The minimum value of \(y = A\sin(x)\) occurs when \(\sin(x)= - 1\). So \(y=2\times(-1)=-2\), and \(x = 270^{\circ}+n\cdot360^{\circ}\). In the given interval \(x = 270^{\circ}\), and the coordinate is \((270^{\circ},-2)\).

Step4: Plot the function

• Start at the origin \((0^{\circ},0)\).
• As \(x\) increases from \(0^{\circ}\) to \(90^{\circ}\), \(y = 2\sin(x)\) increases from \(0\) to \(2\) (reaching the maximum at \((90^{\circ},2)\)).
• As \(x\) increases from \(90^{\circ}\) to \(180^{\circ}\), \(y = 2\sin(x)\) decreases from \(2\) to \(0\) (reaching the x - intercept at \((180^{\circ},0)\)).
• As \(x\) increases from \(180^{\circ}\) to \(270^{\circ}\), \(y = 2\sin(x)\) decreases from \(0\) to \(- 2\) (reaching the minimum at \((270^{\circ},-2)\)).
• As \(x\) increases from \(270^{\circ}\) to \(360^{\circ}\), \(y = 2\sin(x)\) increases from \(-2\) to \(0\) (reaching the x - intercept at \((360^{\circ},0)\)).

Answer:

• x - intercepts: \((0^{\circ},0)\), \((180^{\circ},0)\), \((360^{\circ},0)\)
• Maximum point: \((90^{\circ},2)\)
• Minimum point: \((270^{\circ},-2)\)