The diagram shows part of the graph of a function whose equation is $y = a\sin(bx^{\circ}) + c$
(a) Write down the values of a, b and c.
(b) Determine the exact value of P where the graph intersects the x-axis as shown in the diagram.
For part a, I can say that the amplitude is 2 because it covers 1 to -3. I am confused why d is not in this equation as the graph has been vertically shifted -1 units or is it that c is -1 in this graph?
For B or the period, a period of 1 is $360^{\circ}$ but this graph only goes to 120 so ${360\over120} = 3$,so b is 3.
The phase shift is $-c\over b$ so it is $-1\over3$.
I am not sure how to find part b.
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$\begingroup$a=2, b=3, c=-1
To solve part (b), you need to solve the trigonometric equation $2\sin(3x^\circ)=1$ The solution P is then 50 degrees. (There is another solution at 10 degrees, but that is not P).
$\endgroup$ 0 $\begingroup$The amplitude is indeed $2$, as $\frac{1}{2}(\max(f) - \min(f))$. The average lies at $\frac{1}{2}(\max(f) + \min(f))$ which is $-1$, so the shift downwards is $c = -1$. The amplitude is $a$, so $a = 2$.
There is no phase shift (this would correspond with a number $p$ in $\sin(t-p)$), but there is a type of acceleration, which affects the period. The period is $120^\circ$, instead of $360^\circ$ so there is a threefold increase. The graph is at $120$ degrees where it normally would be at $360$. So $b \cdot 120 = 360$, so $b = 3$.
So in total: $y = 2\sin(3x) -1 $
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