Question

question: consider the function y = 2 sin(x) for 0° ≤ x ≤ 360°. 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: Recall properties 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 is $|A|=2$, the period is $T=\frac{2\pi}{B}=360^{\circ}$ (since $B = 1$ and we are working in degrees).

Step2: Find x - intercepts

Set $y = 0$. Then $2\sin(x)=0$. In the interval $0^{\circ}\leq x\leq360^{\circ}$, $\sin(x)=0$ when $x = 0^{\circ},180^{\circ},360^{\circ}$. The coordinates of the x - intercepts are $(0^{\circ},0),(180^{\circ},0),(360^{\circ},0)$.

Step3: Find maximum points

The maximum value of $\sin(x)$ is 1. For $y = 2\sin(x)$, when $\sin(x)=1$, $y = 2$. In the interval $0^{\circ}\leq x\leq360^{\circ}$, $\sin(x)=1$ when $x = 90^{\circ}$. The coordinate of the maximum - point is $(90^{\circ},2)$.

Step4: Find minimum points

The minimum value of $\sin(x)$ is - 1. For $y = 2\sin(x)$, when $\sin(x)=-1$, $y=-2$. In the interval $0^{\circ}\leq x\leq360^{\circ}$, $\sin(x)=-1$ when $x = 270^{\circ}$. The coordinate of the minimum - point is $(270^{\circ},-2)$.

Step5: Plot the function

Plot the points $(0^{\circ},0),(90^{\circ},2),(180^{\circ},0),(270^{\circ},-2),(360^{\circ},0)$ on the coordinate plane and draw a smooth curve passing through these points. The x - axis is labeled "Angle (degrees)" and the y - axis is labeled "y - value".

Answer:

The key points are $(0^{\circ},0),(90^{\circ},2),(180^{\circ},0),(270^{\circ},-2),(360^{\circ},0)$. Plot these points and draw a smooth sine - curve.