Question

1. Graphs of two trigonometric functions are shown below

(note: not certain if my answers are accurate)

Answer 1

Part VI:
`\displaystyle y = sec\:x`

Part V:
`\displaystyle y = 3csc\:(2x + (\pi)/(2)) \\ y = 3sec\:2x`

Part III: In addition, although this is a secant graph, it can also be a cosecant graph when you apply the C-value of
`\displaystyle -(\pi)/(2)`to the equation.

Part II:
`\displaystyle (\pi)/(2)`

Part I:
`\displaystyle \pi`

Explanation:

`\displaystyle \boxed{y = 3csc\:(2x + (\pi)/(2))} \\ 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 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{-(\pi)/(4)} \hookrightarrow (-(\pi)/(2))/(2) \\ Wavelength\:[Period] \hookrightarrow (\pi)/(2) \\ Amplitude \hookrightarrow N/A`

OR

`\displaystyle \boxed{y = 3sec\:2x} \\ \\ 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 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow (\pi)/(2) \\ Amplitude \hookrightarrow N/A`

Here is all the information you will need. Now, what you need to know is that ALL tangent, secant, cosecant, and cotangent functions have NO amplitudes. Also, keep in mind that although this IS the secant graph, if you plan on writing your equation as a function of cosecant, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of
`\displaystyle y = 3csc\:2x,`in which you need to replase "secant" with "cosecant", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosecant graph [photograph on 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 cosecant graph [photograph on the right] is shifted
`\displaystyle (\pi)/(4)\:unit`to the right, which means that in order to match the secant graph [photograph on the left], we need to shift the graph BACKWARD
`\displaystyle (\pi)/(4)\:unit,`which means the C-term will be negative, and by perfourming your calculations, you will arrive at
`\displaystyle \boxed{-(\pi)/(4)} = (-(\pi)/(2))/(2),`and with that, the cosecant graph of the secant graph, accourding to the horisontal shift, is
`\displaystyle y = 3csc\:(2x + (\pi)/(2)).`Now, with all that being said, we can move forward. To find the period, in this case, you need to take a look at the distanse between each vertical asymptote. Now, accourding to this graph, the vertical asymptotes are at
`\displaystyle (5)/(6)\pi = x, (\pi)/(3) = x, -(\pi)/(6) = x, and\:-(2)/(3)\pi = x.`From here, you will then find the distanse between these asymptotes by simply perfourming the operation of Deduction, and doing that will give you
`\displaystyle \boxed{(\pi)/(2)} = -(\pi)/(6) + (2)/(3)\pi.`So, the period of this function is
`\displaystyle (\pi)/(2).` Now, you will instantly get the jist of the horisontal shift by looking at the above formula. Just keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must pay cloce attention to what is given to you inside those parentheses. Finally, the midline is the centre of the graph, also known as the vertical shift, which in this case is at
`\displaystyle y = 0.`

*So, double-check your first two answers because they are incorrect.

I am delighted to assist you at any time.