Write a sinsoidal equation that has: maximum points (π/6, 14) (π, 14) and minimum points (-π/4, -10) (7π/12, -10)

Question

I was able to figure out that the amplitude is 12 and the vertical shift is 2 but I am struggling to figure out the period and the phase shift.

Answer 1

y = Asin(B(x+C))+D

You already determined that A = 12 and D = 2

For max: B(π/6+C) = π/2 --> sin(π/2) = max of 1

For min: B(-π/4+C) = -π/2 -->sin(-π/2) = min of -1

Solving the simultaneous equations

B = 12/5 and C = π/24

y = 12sin(12/5(x+π/24)) + 2

Checking:

y(π/6) = 12sin(π/2)+ 2 = 14

y(π) = 12sin(5π/2) + 2 = 14

y(-π/4) = 12sin(-5π/2) = -12+2 = -10

y(7π/12)=12sin(3π/2) = -12+2 = -10

Answer 2

Since the question only specifies a "sinusoidal", either a sine function or a cosine function would work, but let's stick with the true "sinusoid" and pick a sine function. The period is the "x" distance from min to min or from max to max. So using the maximum points the period is (π - π/6) = 6π/6 - π/6 = 5π/6. Checking the minimum points for concurrence, we get (7π/12 - -π/4) = 7π/12 + 3π/12 = 10π/12 = 5π/6.

The generic equation is y = Asin(B(x - C)) + D where A is the amplitude, B is 2π/period, C is the horizontal shift, and D is the vertical shift.

A = [14 - (-10)]/2 = 12

B = (2π)/(5π/6) = (2π)/1 • 6/(5π) = 12/5

D = 2 as you mention

To calculate "C", normally the sine wave starts at (0, 0) but in this case (2, 0). But in this case to find where it starts, we go halfway between the minimum of -π/4 and the max at π/6 so (-π/4 + π/6)/2 = (-3π/12 + 2π/12)/2 = (-π/12)/2 = -π/24

So y = 12sin(12/5(x + π/24)) + 2

If you want you can simplify a bit by distributing the 12/5 to get y = 12sin(12/5x + π/10)

Answer 3

Note that the first max - first min, second min - first max, and second max - first min are all 5π/12. This means that our calculation of period will be straightforward (the maximums will be 1 period apart, the minimums will be one period apart). Thus, the period P = π - π/6 =5π/6. ω =2π/P = 12/5

We can pick two locations to calculate the phase shift from (min followed by the max locates phase shift in the middle, as this is where 0 is for the sin function), but it makes most sense to include one containing 0 in the interval. Then, (-π/4 + π/6)/2 = -π/24

Thus, φ = -π/24 (Note that -π/4 + π/24 = -5π/24 and π/6 + π/24 = 5π/24)

Answer 4

To find the period, Subtract the max values, 6π/6-π/6 = 5π/6 For finding the b value for the equation, make sure to do the following:

2π/b = 5π/6 and solve for b.

For the phase shift, the easiest to do is use a cos equation, then it would be x-π/6. Sin would be a little more difficult, you would need to find the x value where it crosses the x axis.

Answer 5

f(x) = Asin(Bx + C) + D;

A = (14 - (-10))/2 = 12; DE = (14 + (- 10))/2 = 2.

So we have now f(x) = 12sin(Bx + C) + 2;

x = π/6, B⋅π/6 + C = π/2;

x = - π/4, B⋅(- π/4) + C = 3π/2

If sabtruct these equations we get: 5π/12⋅B = - π; B = -12/5;

C = π/2 + 12/5⋅π/6 = π/2 + 2π/5 = 9π/10.

So, f(x) = 12sin(- 12/5x + 9π/10) + 2 = 12sin(12/5x + ℵ/10)