Answer: A sine graph \(y=2\sin(x/3)-1\), crossing \((0,-1)\), with minimum \((-270^\circ,-3)\), maximum \((270^\circ,1)\), and crossing the \(x\)-axis at \((90^\circ,0)\).
For a trigonometric graph, identify the amplitude, midline and period first, then use these to locate the key points of the sketch.
The graph is a transformed sine curve with centre line \(y=-1\), amplitude \(2\), and angle input \(\frac{x}{3}\).
Useful points are:
\(x=-270^\circ:\quad y=2\sin(-90^\circ)-1=-3,\)
\(x=0^\circ:\quad y=2\sin0^\circ-1=-1,\)
\(x=90^\circ:\quad y=2\sin30^\circ-1=0,\)
\(x=270^\circ:\quad y=2\sin90^\circ-1=1.\)
So the curve starts in the lower half of the axes, passes through \((0,-1)\), crosses the \(x\)-axis at \(90^\circ\), and reaches a maximum at \((270^\circ,1)\).
Therefore the result matches the required answer.
