Question

3. Based on the graph of the trigonometric function,

what is the period?

4. Based on the graph of the trigonometric function,
what is the amplitude?

Answer 1

4. ☐ A
`\displaystyle (1)/(3)`

☐ B
`\displaystyle 2`

☑ C
`\displaystyle 3`

3. ☐ A
`\displaystyle (\pi)/(3)`

☐ B
`\displaystyle \pi`

☑ C
`\displaystyle 6\pi`

Explanation:

`\displaystyle y = -2sin\:((1)/(3)x + (\pi)/(2)) \\ y = -2cos\:(1)/(3)x`

`\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \hookrightarrow \boxed{-1(1)/(2)\pi} \hookrightarrow (-(\pi)/(2))/((1)/(3)) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{6\pi} \hookrightarrow (2)/((1)/(3))\pi \\ Amplitude \hookrightarrow 2`

OR

`\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow (C)/(B) \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow (2)/(B)\pi \hookrightarrow \boxed{6\pi} \hookrightarrow (2)/((1)/(3))\pi \\ Amplitude \hookrightarrow 2`

The above information can help you interpret the graph much better. First off, JUST IN CASE you needed to know the trigonometric equation(s) of this graph, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, 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 = -2sin\:(1)/(3)x,` in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine 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 sine graph [photograph on the right] is shifted
`\displaystyle 1(1)/(2)\pi\:unit` to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACK
`\displaystyle 1(1)/(2)\pi\:unit,` which means the C-term will be negative, and perfourming your calculations, you will arrive at
`\displaystyle \boxed{-1(1)/(2)\pi} = (-(\pi)/(2))/((1)/(3)).` So, the sine graph of the cosine graph, accourding to the horisontal shift, is
`\displaystyle y = -2sin\:((1)/(3)x + (\pi)/(2)).` Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits
`\displaystyle [0, -2],` from there to
`\displaystyle [-6\pi, -2],` they are obviously
`\displaystyle 6\pi\:units` apart, telling you that the period of the graph is
`\displaystyle 6\pi.` Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at
`\displaystyle y = 0,` in which each crest is extended two units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow. Now, in this case, you were probably wondering why the negative was inserted in front of the amplitude in the equation(s). Well, here is why:

`\displaystyle y = sin\:x \\ y = cos\:x`

Knowing your parent functions, sine commenses upright from the origin, while cosine commenses one unit above the origin
`\displaystyle [0, 1].` This should tell you that inserting a negative in front of the amplitude will REFLECT each crest over the midline. Well, there you have it.

I am delighted to assist you at any time.