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 the mid - line is $y = 0$.

Step2: Find x - intercepts

Set $y = 0$. So, $2\sin(x)=0$. 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 $(0,0),(180,0),(360,0)$.

Step3: Find maximum points

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

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 $(270,-2)$.

Step5: Plot the function

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 connect them with a smooth curve to get the graph of $y = 2\sin(x)$ in the interval $0^{\circ}\leq x\leq360^{\circ}$.

Answer:

The x - intercepts are $(0,0),(180,0),(360,0)$, the maximum point is $(90,2)$, and the minimum point is $(270,-2)$. Plot these points and connect them with a smooth curve to graph $y = 2\sin(x)$ for $0^{\circ}\leq x\leq360^{\circ}$.