Christoforos A.
asked • 06/16/21How 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
2 Answers By Expert Tutors
David W. answered • 06/16/21
Advanced Calculus, Vector Analysis, Diff Eq ... Statistics in college
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
- amplitude is A
- period is 2π/B
- phase shift is C (positive is to the left)
- vertical shift is D
Andrew D. answered • 06/16/21
Bachelor's degree in mathematics with coursework in discrete math
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.
Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Christoforos A.
I forgot to add on, as I plotted this equation graph went past π. Where did I go wrong and how can I fix it?06/16/21