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The graph of a sinusoidal function has a minimum point at (0,-3) and then intersects its midline at (1,1). Write the formula of the function, where x is entered in radians. f(x)=

The graph of a sinusoidal function has a minimum point at (0,−3) (0,-3) and then intersects its midline at (1,1) (1,1) .\newlineWrite the formula of the function, where x x is entered in radians.\newlinef(x)= f(x)=

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Q. The graph of a sinusoidal function has a minimum point at (0,−3) (0,-3) and then intersects its midline at (1,1) (1,1) .\newlineWrite the formula of the function, where x x is entered in radians.\newlinef(x)= f(x)=
  1. Identify Amplitude: Since the graph has a minimum point at (0,−3)(0, -3), we know that this is the lowest point of the function. For a sinusoidal function such as a sine or cosine function, the minimum point occurs at the amplitude below the midline. Therefore, the amplitude of the function is the distance from the midline to the minimum point.
  2. Calculate Amplitude: The midline of the function is at y=1y = 1, and the minimum point is at y=−3y = -3. The amplitude (AA) is the distance from the midline to the minimum, which is 1−(−3)=41 - (-3) = 4.
  3. Determine Function Form: Since the minimum point is at (0,−3)(0, -3), we can use a cosine function that starts at its minimum value when x=0x = 0. This means the cosine function will be reflected over the xx-axis. The general form of the function will be f(x)=−Acos⁡(Bx)+Df(x) = -A\cos(Bx) + D, where DD is the midline.
  4. Find Period: We have found that A=4A = 4 and D=1D = 1. Now we need to determine the value of BB, which affects the period of the function. The period (T)(T) of a cosine function is given by T=2πBT = \frac{2\pi}{B}. Since the function intersects its midline at (1,1)(1, 1), this is a quarter of the period. Therefore, the full period is 44 times this value, which is 4×1=44 \times 1 = 4.
  5. Calculate B Value: To find B, we use the period formula T=2πBT = \frac{2\pi}{B}. We have T=4T = 4, so B=2πT=2π4=π2B = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}.
  6. Final Equation: Substituting A=4A = 4, B=π2B = \frac{\pi}{2}, and D=1D = 1 into the equation f(x)=−Acos⁡(Bx)+Df(x) = -A\cos(Bx) + D gives us the final equation f(x)=−4cos⁡(π2⋅x)+1f(x) = -4\cos(\frac{\pi}{2} \cdot x) + 1.

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