Question

consider the function $y = 2sin(x)$ for $0^{circ}leq xleq360^{circ}$. 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 a 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 $2\sin(x)=0$. So, $\sin(x)=0$. In the interval $0^{\circ}\leq x\leq360^{\circ}$, $x = 0^{\circ},180^{\circ},360^{\circ}$. The coordinates of the x - intercepts are $(0,0)$, $(180,0)$, $(360,0)$.

Step3: Find maximum points

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

Step4: Find minimum points

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

Step5: Plot the points and draw the graph

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

Answer:

The x - intercepts are $(0,0)$, $(180,0)$, $(360,0)$; the maximum - point is $(90,2)$; the minimum - point is $(270,-2)$. Plot these points and draw a sine - curve with amplitude 2 and period $360^{\circ}$ on the coordinate plane with appropriate axis labels.