Question

question:
consider the function ( y = 2sin(x) ) for ( 0^{circ}leq xleq360^{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: Identify the general form of sine function

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

Step2: Find x - intercepts

Set \(y = 0\), so \(2\sin(x)=0\). Then \(\sin(x)=0\). For \(0^{\circ}\leq x\leq360^{\circ}\), \(x = 0^{\circ},180^{\circ},360^{\circ}\). The coordinates are \((0^{\circ},0)\), \((180^{\circ},0)\), \((360^{\circ},0)\).

Step3: Find maximum points

The maximum value of \(y = A\sin(x)\) occurs when \(\sin(x)=1\). Since \(A = 2\), when \(\sin(x)=1\) (i.e., \(x = 90^{\circ}\)), \(y=2\times1 = 2\). The coordinate is \((90^{\circ},2)\).

Step4: Find minimum points

The minimum value of \(y = A\sin(x)\) occurs when \(\sin(x)=- 1\). Since \(A = 2\), when \(\sin(x)=-1\) (i.e., \(x = 270^{\circ}\)), \(y=2\times(-1)=-2\). The coordinate is \((270^{\circ},-2)\).

To graph the function:

• Start at the origin \((0^{\circ},0)\).
• Rise to the maximum point \((90^{\circ},2)\).
• Then fall back to the x - axis at \((180^{\circ},0)\).
• Continue to the minimum point \((270^{\circ},-2)\).
• Finally, rise back to the x - axis at \((360^{\circ},0)\).

Answer:

Key points:

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