Question

What is the period and midline?

Answer 1

$\bf \qquad \qquad \qquad \qquad \textit{function transformations} \\ \quad \\ % function transformations for trigonometric functions \begin{array}{rllll} % left side templates f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}} \\ \quad \\ \end{array}\qquad$

$\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks}\\ \quad \textit{horizontally by amplitude } |{{ A}}|\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\ \bullet \textit{vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\ \end{array}$

$\bf \begin{array}{llll} \bullet \textit{function period}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}$

so if you notice yours $\bf \begin{array}{llll} 3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\ &\ \uparrow&\uparrow \\ &B&D \end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
 has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline  up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)