Question

Explain how the graph of y = 2csc(x - pi/4) - 5 is related to the graph of the basic trigonometric function y = cscx.

Answer 1

Step-by-step explanation:

`\displaystyle \boxed{y = 2sec\:(x - (3)/(4)\pi) - 5} \\ y = Asec(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (\pi)/(B) \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -5 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{(3)/(4)\pi} \hookrightarrow ((3)/(4)\pi)/(1) \\ Wavelength\:[Period] \hookrightarrow (\pi)/(B) \hookrightarrow \boxed{\pi} \hookrightarrow (\pi)/(1) \\ Amplitude \hookrightarrow N/A`

OR

`\displaystyle y = Acsc(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (\pi)/(B) \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow -5 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{(\pi)/(4)} \hookrightarrow ((\pi)/(4))/(1) \\ Wavelength\:[Period] \hookrightarrow (\pi)/(B) \hookrightarrow \boxed{\pi} \hookrightarrow (\pi)/(1) \\ Amplitude \hookrightarrow N/A`

Now, what you need to know is that ALL tangent, secant, cosecant, and cotangent functions have NO amplitudes. Also, keep in mind that although you are given a cosecant equation, if you plan on writing your equation as a function of secant, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph farthest to the right displays the trigonometric graph of
`\displaystyle y = 2sec\:(x - (\pi)/(4)) - 5,` in which you need to replase "cosecant" with "secant", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosecant graph [photograph farthest to the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the secant graph [photograph farthest to the right] is shifted
`\displaystyle (\pi)/(2)\:unit`to the left, which means that in order to match the cosecant graph [photograph farthest to the left], we need to shift the graph FORWARD
`\displaystyle (\pi)/(2)\:unit,`for a total of
`\displaystyle (3)/(4)\pi\:units\:[(\pi)/(4) + (\pi)/(2)],`which means the C-term will be postive, and by perfourming your calculations, you will arrive at
`\displaystyle \boxed{(3)/(4)\pi} = ((3)/(4)\pi)/(1).`So, the secant equation of the cosecant equation, accourding to the horisontal shift, is
`\displaystyle y = 2sec\:(x - (3)/(4)\pi) - 5.`Now, with all that being said, in this case, sinse you ONLY have a wourd problem to wourk with, you MUST use the above formula for how to calculate the period. Onse you figure this out, the rest should be simple. Now, the midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at
`\displaystyle y = -5,`in which the graph will have a vertical-stretch factour of two units beyond the midline. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

**The equation of
`\displaystyle y = csc\:x`is represented by the centre photograph.

I am delighted to assist you at any time.

Answer 2

Answer: The distance between the min and max increased by a factor of 2, horizontally shifted to the right by
`\bold{(\pi)/(4)}`, shifted down 5 units, and has a phase shift of
`\bold{(\pi)/(4)}`

Step-by-step explanation:

The general form of a csc function is: y = A csc (Bx - C) + D where

• |A| is the amplitude (vertical stretch)
• -A is a reflection over the x-axis
• |B| is the horizontal stretch
• -B is a reflection over the y-axis
• C is a horizontal shift (left or right)
• D is a vertical shift (up or down)

• Period is
  `(2\pi)/(B)`
• Phase Shift is
  `(C)/(B)`

In the given transformed graph y = 2 csc (x -
`(\pi)/(4)`) - 5

• A = 2 --> distance between the Min and max is 2(2) = 4
• C =
  `(\pi)/(4)` --> horizontally shifted to the right
  `(\pi)/(4)` units
• D = -5 --> shifted DOWN 5 units
• Phase shift is
  `(\pi)/(4)`