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consider the function ( y = 2sin(x) ) for ( 0^{circ}leq xleq360^{circ} …

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: Find x - axis intersection points

Set \(y = 0\), then \(2\sin(x)=0\), \(\sin(x) = 0\).
For \(0^{\circ}\leq x\leq360^{\circ}\), \(x = 0^{\circ},180^{\circ},360^{\circ}\).

Step2: Find maximum and minimum points

The general form of \(y = A\sin(x)\) has amplitude \(|A|\). Here \(A = 2\).
The maximum value of \(y=\sin(x)\) is \(1\) when \(x = 90^{\circ}\), so for \(y = 2\sin(x)\), \(y=2\) at \(x = 90^{\circ}\).
The minimum value of \(y=\sin(x)\) is \(- 1\) when \(x = 270^{\circ}\), so for \(y = 2\sin(x)\), \(y=-2\) at \(x = 270^{\circ}\).

Answer:

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