8. assume this is a sine graph. the point $(-2\\pi,0)$ should be considered as the transformed starting…

8. 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) $140^{\\circ}$

8. 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) $140^{\\circ}$

Answer

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})