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.
Best Answer
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).