Question

1. assume this is a sine graph. the point $(-2\pi,0)$ should be considered as the transformed starting point for $y = \sin(x)$. phase shift: $h = $ vertical shift: $k = $ period: $b = $ amplitude: $a = $ $y = $ 9. a) assume this is a sine graph. the point $(0,-2)$ should be considered as the transformed starting point for $y = \sin(x)$. phase shift: $h = $ vertical shift: $k = $ period: $b = $ amplitude: $a = $ $y = $ b) now assume that this is a cosine graph. the point $(\frac{\pi}{4},3)$ should be considered as the transformed starting point for $y = \cos(x)$. phase shift: $h = $ $y = $ 10. convert between radians and degrees. give exact answers. a) $-\frac{7\pi}{10}$ b) $140^{\circ}$

Explanation:

Step1: Analyze the general form of the sine function

The general form of a sine function is \(y = a\sin(b(x - h))+k\), where \(a\) is the amplitude, \(b\) is related to the period (\(T=\frac{2\pi}{b}\)), \(h\) is the phase - shift, and \(k\) is the vertical shift.

Step2: For problem 8

• Phase - shift:

The starting point of \(y = \sin(x)\) is \((0,0)\). Here, the starting point is \((- 2\pi,0)\). Using the formula \(x=h\) for the starting point of \(y = \sin(b(x - h))+k\), we have \(h=-2\pi\). The phase - shift is \(2\pi\) to the left.

• Vertical shift:

The mid - line of the graph. The maximum value is \(y = 2\) and the minimum value is \(y=-2\). The mid - line \(y = k\). Using the formula \(k=\frac{y_{max}+y_{min}}{2}=\frac{2+( - 2)}{2}=0\)

• Period:

The distance between two consecutive similar points (e.g., two maxima). From the graph, the period \(T = 8\pi\). Using \(T=\frac{2\pi}{b}\), we solve \(8\pi=\frac{2\pi}{b}\), so \(b=\frac{1}{4}\)

• Amplitude:

The distance from the mid - line to the maximum (or minimum) value. \(a = 2\)

• Function:

Substituting \(a = 2\), \(b=\frac{1}{4}\), \(h=-2\pi\), \(k = 0\) into \(y=a\sin(b(x - h))+k\), we get \(y = 2\sin(\frac{1}{4}(x + 2\pi))\)

Step3: For problem 9a

• Phase - shift:

The starting point of \(y=\sin(x)\) is \((0,0)\). Here, the starting point is \((0,-2)\). For \(y=a\sin(b(x - h))+k\), \(h = 0\). The phase - shift is \(0\)

• Vertical shift:

The mid - line. Let's find two points. The maximum value (from the general shape, assume a standard period - related calculation). Using \(k=-2\) (since the starting point has \(y\) - value \(-2\) and if we assume no phase - shift in the \(x\) direction for the starting point in the \(x\) - axis sense for the transformed starting point given as \((0,-2)\))

• Period:

Assume a standard period (by looking at the distance between two similar points). If we assume the period \(T=\pi\). Using \(T=\frac{2\pi}{b}\), then \(b = 2\)

• Amplitude:

The distance from the mid - line (\(y=-2\)) to a maximum (e.g., if we assume a point above). Let's say the maximum is \(y = 1\) (from the graph's general trend, assume a value relative to the vertical shift). \(a=3\)

• Function:

Substituting \(a = 3\), \(b = 2\), \(h = 0\), \(k=-2\) into \(y=a\sin(b(x - h))+k\), we get \(y = 3\sin(2x)-2\)

Step4: For problem 9b (cosine function)

The general form of a cosine function is \(y=a\cos(b(x - h))+k\)

• Phase - shift:

The starting point of \(y = \cos(x)\) is \((0,1)\). Here, the starting point is \((\frac{\pi}{4},3)\). So \(h=\frac{\pi}{4}\). The phase - shift is \(\frac{\pi}{4}\) to the right

• Vertical shift:

Using \(k = 3\) (the \(y\) - value of the starting point for the transformed cosine function)

• Assume period (if we assume a standard - looking graph):

If we assume \(T=\pi\), then \(b = 2\) (from \(T=\frac{2\pi}{b}\))

• Amplitude:

Assume \(a = 0\) (this part may need more graph details, but if we assume the function is \(y=\cos(2(x-\frac{\pi}{4}))+3\))

Step5: For problem 10a (radians to degrees)

Use the conversion formula \(x\) radians \(=x\times\frac{180^{\circ}}{\pi}\)
For \(x=-\frac{7\pi}{10}\), \(y=-\frac{7\pi}{10}\times\frac{180^{\circ}}{\pi}=-126^{\circ}\)

Step6: For problem 10b (degrees to radians)

Use the conversion formula \(x\) degrees \(=x\times\frac{\pi}{180}\)
For \(x = 140^{\circ}\), \(y=140\times\frac{\pi}{180}=\frac{7\pi}{9}\)

Answer:

• Problem 8:
• Phase shift: \(2\pi\) to the left, \(h=-2\pi\)
• Vertical shift: \(0\), \(k = 0\)
• Period: \(8\pi\), \(b=\frac{1}{4}\)
• Amplitude: \(2\), \(a = 2\)
• \(y = 2\sin(\frac{1}{4}(x + 2\pi))\)
• Problem 9a:
• Phase shift: \(0\), \(h = 0\)
• Vertical shift: \(-2\), \(k=-2\)
• Period: \(\pi\) (assumed, adjust if more graph details), \(b = 2\)
• Amplitude: \(3\) (assumed, adjust if more graph details), \(a = 3\)
• \(y = 3\sin(2x)-2\)
• Problem 9b:
• Phase shift: \(\frac{\pi}{4}\) to the right, \(h=\frac{\pi}{4}\)
• \(y=\cos(2(x-\frac{\pi}{4}))+3\) (assuming \(b = 2\) from period assumption)
• Problem 10a: \(-126^{\circ}\)
• Problem 10b: \(\frac{7\pi}{9}\)