Math Problem Statement

Determine the general sine function, 𝑓=𝐴+𝐡sin(𝐢(π‘₯+𝐷)) that has a minimum value of βˆ’47 at π‘₯=6 , a maximum value of 153 and a period of 76 .

Solution

To determine the general sine function f(x)=A+Bsin⁑(C(x+D))f(x) = A + B \sin(C(x + D)) that satisfies the given conditions, we need to find the values of AA, BB, CC, and DD. Let's break down the conditions and derive the values step by step.

Step 1: Amplitude and Vertical Shift

The function has a minimum value of -47 and a maximum value of 153. The sine function oscillates between its minimum and maximum values, so:

  • The amplitude BB is half the distance between the maximum and minimum values: B=153βˆ’(βˆ’47)2=2002=100B = \frac{153 - (-47)}{2} = \frac{200}{2} = 100

  • The midline AA is the average of the maximum and minimum values, which represents the vertical shift: A=153+(βˆ’47)2=1062=53A = \frac{153 + (-47)}{2} = \frac{106}{2} = 53

So far, we have A=53A = 53 and B=100B = 100.

Step 2: Period and Frequency

The period of a sine function is related to the coefficient CC. The period of a standard sine function sin⁑(Cx)\sin(Cx) is 2Ο€C\frac{2\pi}{C}. We are given that the period is 76, so we can set up the equation: 2Ο€C=76\frac{2\pi}{C} = 76 Solving for CC: C=2Ο€76=Ο€38C = \frac{2\pi}{76} = \frac{\pi}{38}

Step 3: Phase Shift

The function has a minimum value of -47 at x=6x = 6. The sine function sin⁑(x)\sin(x) typically reaches its minimum at sin⁑(3Ο€2)=βˆ’1\sin\left(\frac{3\pi}{2}\right) = -1. Therefore, the phase shift DD ensures that the sine function reaches this minimum at x=6x = 6.

We want the argument of the sine function to equal 3Ο€2\frac{3\pi}{2} when x=6x = 6: C(x+D)=3Ο€2whenx=6C(x + D) = \frac{3\pi}{2} \quad \text{when} \quad x = 6 Substitute C=Ο€38C = \frac{\pi}{38}: Ο€38(6+D)=3Ο€2\frac{\pi}{38}(6 + D) = \frac{3\pi}{2} Solving for DD: 6+D=3Ο€2Γ—38Ο€=576 + D = \frac{3\pi}{2} \times \frac{38}{\pi} = 57 D=57βˆ’6=51D = 57 - 6 = 51

Final Equation

Now that we have all the parameters, the general sine function is: f(x)=53+100sin⁑(Ο€38(x+51))f(x) = 53 + 100 \sin\left(\frac{\pi}{38}(x + 51)\right)

This is the sine function that satisfies the given conditions.

Would you like more details on any of the steps? Here are five related questions to expand your understanding:

  1. How does changing the amplitude affect the graph of a sine function?
  2. What is the role of the phase shift in a sine function?
  3. How would the function change if the period were doubled?
  4. Can you describe how the vertical shift AA modifies the graph?
  5. How can you generalize the approach to finding sine functions with different extrema?

Tip: The value CC controls the period of the sine wave, so always use the relation 2Ο€C=period\frac{2\pi}{C} = \text{period} when solving for CC.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Function
Amplitude
Period
Phase Shift
Vertical Shift

Formulas

General sine function: f(x) = A + B sin(C(x + D))
Amplitude: B = (max - min) / 2
Vertical Shift: A = (max + min) / 2
Period relation: Period = 2Ο€ / C
Phase Shift: C(x + D) = 3Ο€ / 2 when x is at the minimum

Theorems

Period of sine functions
Amplitude and vertical shift properties of trigonometric functions

Suitable Grade Level

Grades 11-12