Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The graph of a sinusoidal function has a maximum point at 
(0,8) and then has a minimum point at 
(5,2).
Write the formula of the function, where 
x is entered in radians.

f(x)=____

The graph of a sinusoidal function has a maximum point at (0,8) (0,8) and then has a minimum point at (5,2) (5,2) .\newlineWrite the formula of the function, where x x is entered in radians.\newline f(x)=_____

Full solution

Q. The graph of a sinusoidal function has a maximum point at (0,8) (0,8) and then has a minimum point at (5,2) (5,2) .\newlineWrite the formula of the function, where x x is entered in radians.\newline f(x)=_____
  1. Calculate Amplitude: Determine the amplitude of the sinusoidal function. The amplitude is half the distance between the maximum and minimum values of the function. Here, the maximum value is 88 and the minimum value is 22. So, the amplitude AA is calculated as:\newlineA=822=3.A = \frac{8 - 2}{2} = 3.
  2. Identify Vertical Shift: Identify the vertical shift of the function. The vertical shift is the average of the maximum and minimum values. Calculate it as:\newlineVertical shift = (8+2)/2=5(8 + 2) / 2 = 5.
  3. Calculate Period: Calculate the period of the function. Since the function goes from a maximum at x=0x = 0 to the next minimum at x=5x = 5, half the period is 55. Therefore, the full period TT is: T=5×2=10T = 5 \times 2 = 10.
  4. Determine Horizontal Shift: Determine the horizontal shift (phase shift) of the function. Since the maximum occurs at x=0x = 0, there is no horizontal shift. Thus, the phase shift cc is 00.
  5. Write Sinusoidal Function Formula: Write the formula of the sinusoidal function using the values calculated. The general form of a sinusoidal function is:\newlinef(x)=Asin(B(xC))+Df(x) = A \cdot \sin(B(x - C)) + D\newlineWhere AA is the amplitude, BB is the frequency (B=2πTB = \frac{2\pi}{T}), CC is the phase shift, and DD is the vertical shift. Plugging in the values:\newlinef(x)=3sin(2π10(x0))+5f(x) = 3 \cdot \sin\left(\frac{2\pi}{10}(x - 0)\right) + 5\newlineSimplify BB:\newlinef(x)=3sin(π5x)+5.f(x) = 3 \cdot \sin\left(\frac{\pi}{5} \cdot x\right) + 5.

More problems from Coterminal and reference angles