How would I write the equation of this graph? It should be written in y=ASin(Bx-C)+D I wrote Y=3sin(2x+ π/4)+1

Question

Answer 1

Compare this graph to a sine plot !!

y values range from -2 to 4 (a range of 6; an amplitude of 3). A typical sine plot ranges from -1 to +1 (an amplitude of 1), so this is 3 times as much and y is +1 more. That makes A=3 and D=1.

y = 3 sin(Bx-C) + 1 [so far]

Note that the period of a sine plot is from 0 to 2π and the sin(x)=0 at x=0, x=π and x=2π . Here, the x-values are 1/2 of those (see red line in my figure), so they must be multiplied by B=2 They are not offset, so C=0)

The result is:

y = 3 sin(2x) + 1

y = A sin(B(x + C)) + D

1. amplitude is A
2. period is 2π/B
3. phase shift is C (positive is to the left)
4. vertical shift is D

Answer 2

In this case, you are mostly correct. The amplitude is 3, which you indicated by the 3 in front. The function has been shifted up by 1, which you also did correctly. And the period is pi, which is also correctly noted by your inclusion of 2x. The only issue is that you indicated that there was a horizontal shift left by pi/4, but since this is a sine graph, then there is actually no horizontal shift. So, you should simply exclude that pi/4 term.