Question

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

Answer 1

Answer:

The sinusoidal function is $y = 1 + 4 \cdot \sin \left[\frac{\pi}{2} \cdot \left(x-1\right)\right]$.

Step-by-step explanation:

The sinusoidal function has the following form:

$y = x_{o} + A \cdot \sin (\omega \cdot x + \phi)$

Where:

$A$ - Amplitude, dimensionless.

$\omega$ - Angular frequency, measured in radians.

$x_{o}$ - Independent component of the midpoint value, dimensionless.

$\phi$ - Phase angle, measured in radians.

Amplitude is the absolute value of the difference between dependent component of the midline and absolute minimum:

$A = |1-(-3)|$

$A = 4$

Let suppose that sinusoidal function has its initial value at absolute minimum. Then:

$x_{o} = 1$

Now, phase angle has to be determined:

$-3 = 1 + 4\cdot \sin (\omega \cdot 0 + \phi)$

$-4 = 4 \cdot \sin \phi$

$\sin \phi = -1$

$\phi = \sin^{-1}\,(-1)$

$\phi = -\frac{\pi}{2}$

The angular frequency of the function is now determined by substituting all remaining variables and clearing it within sinusoidal function:

$1 = 1 + 4\cdot \sin \left(\omega \cdot 1 - \frac{\pi}{2} \right)$

$4\cdot \sin \left(\omega - \frac{\pi}{2} \right) = 0$

$\sin \left(\omega - \frac{\pi}{2} \right) = 0$

$\omega - \frac{\pi}{2} = \sin^{-1}\,0$

$\omega = \frac{\pi}{2}$

Lastly, the sinusoidal function is:

$y = 1 + 4\cdot \sin \left(\frac{\pi}{2} \cdot x - \frac{\pi}{2} \right)$

$y = 1 + 4 \cdot \sin \left[\frac{\pi}{2} \cdot \left(x-1\right)\right]$

Answer 2

Answer:f(x)=−4cos( π/2 x)+1

Step-by-step explanation: