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0606 P23 - Jun 2019 - Q4 - 5 marks
8312
The function \(f\) is defined, for \(0^\circ\leq x\leq360^\circ\), by \(f(x)=4+3\sin2x\).
(i) Sketch the graph of \(y=f(x)\).
(ii) State the period of \(f\).
(iii) State the amplitude of \(f\).
Solution
Answer: Period \(180^\circ\); amplitude \(3\).
The graph is a sine graph with midline \(y=4\), amplitude \(3\), and period \(180^\circ\).
It starts at \(y=4\) when \(x=0^\circ\), reaches a maximum of \(7\) at \(x=45^\circ\), returns to \(4\) at \(x=90^\circ\), reaches a minimum of \(1\) at \(x=135^\circ\), and returns to \(4\) at \(x=180^\circ\). The same cycle repeats from \(180^\circ\) to \(360^\circ\).
Since the coefficient of \(x\) inside the sine is 2, the period is
\(\frac{360^\circ}{2}=180^\circ.\)
Thus
\(\boxed{\text{period}=180^\circ}.\)
The amplitude is the coefficient of the sine term, so
