consider the graph
6
4
2
What is the midline? urte as an equation)
What is the period of the function?
What is the amplitude of the function?
What is the equation of the function? f(x)=
Help please

Answer:

[tex]\displaystyle \boxed{y = -3} \\ \boxed{\pi} \\ \boxed{N/A} \\ f(x) = \boxed{-2cot\:(x + \frac{\pi}{2}) - 3}[/tex]

Step-by-step explanation:

[tex]\displaystyle f(x) = Acot(Bx - C) + D \\ \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{\pi}{B} \\ Amplitude \hookrightarrow |A| \\ \\ \\ Vertical\:Shift \hookrightarrow -3 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{2}} \hookrightarrow \frac{-\frac{\pi}{2}}{1} \\ Wavelength\:[Period] \hookrightarrow \frac{\pi}{2} \\ Amplitude \hookrightarrow N/A[/tex]

OR

[tex]\displaystyle f(x) = \boxed{2tan\:x - 3} \\ \\ f(x) = Atan(Bx - C) + D \\ \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{\pi}{B} \\ Amplitude \hookrightarrow |A| \\ \\ \\ Vertical\:Shift \hookrightarrow -3 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{\pi}{2} \\ Amplitude \hookrightarrow N/A[/tex]

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 a tangent graph, if you plan on writing your equation as a function of cotangent, 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 [tex]\displaystyle f(x) = 2cot\:x - 3,[/tex]in which you need to replase "tangent" with "cotangent", then figure the appropriate C-term out that will make the graph horisontally shift and map onto the tangent 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 graphs, we can tell that the cotangent graph [photograph on the right] is shifted [tex]\displaystyle \frac{\pi}{2}\:unit[/tex]to the right, which means that in order to match the tangent graph [photograph on the left], we need to shift the graph BACKWARD [tex]\displaystyle \frac{\pi}{2}\:unit,[/tex]which means the C-term will be negative, and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{-\frac{\pi}{2}} = \frac{-\frac{\pi}{2}}{1},[/tex]but we are NOT YET DONE. Although we shifted the graph back into position, remember, cotangent is graphed in REVERCE, so we need to insert a negative in front of the amplitude term, and with that, the cotangent graph of the tangent graph, accourding to the horisontal shift, is [tex]\displaystyle f(x) = -2cot\:(x + \frac{\pi}{2}) - 3.[/tex]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 [tex]\displaystyle -\frac{\pi}{2} = x\:and\:\frac{\pi}{2} = x.[/tex]From here, you will then find the distanse between these asymptotes by simply perfourming the operation of Deduction, and doing that will give you [tex]\displaystyle \boxed{\pi} = \frac{\pi}{2} + \frac{\pi}{2}.[/tex]So, the period of this function is [tex]\displaystyle \pi.[/tex]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 your graph, also known as the vertical shift, which in this case is at [tex]\displaystyle y = -3.[/tex]

Well, that just about wraps it up. Now that everything has been explained, you should understand it better. If you still have questions, do not hesitate to comment any you have.

I am delighted to assist you at any time.

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