Answer
$$\eqalign{
& {\text{amplitude: 3}}{\text{.6}} \cr
& {\text{period: 48}} \cr} $$
Work Step by Step
$$\eqalign{
& q\left( x \right) = 3.6\cos \left( {\frac{{\pi x}}{{24}}} \right) \cr
& {\text{The functions of the form }}y = A\cos \left( {B\left( {\theta - C} \right)} \right) + D{\text{ have a vertical}} \cr
& {\text{stretch }}\left( {{\text{or }}{\bf{amplitude}}} \right){\text{ of }}\left| A \right|,{\text{ and a period of }}\frac{{2\pi }}{{\left| B \right|}} \cr
& {\text{Therefore, rewriting the function }}q\left( x \right) = 3.6\cos \left( {\frac{{\pi x}}{{24}}} \right){\text{ we obtain}} \cr
& \underbrace {q\left( x \right) = 3.6\cos \left( {\frac{\pi }{{24}}\left( {x - 0} \right)} \right) + 0}_{y = A\cos \left( {B\left( {\theta - C} \right)} \right) + D} \cr
& A = 3.6,{\text{ }}B = \frac{\pi }{{24}},{\text{ }}C = 0,{\text{ }}D = 0 \cr
& {\text{The amplitude is: }}\left| A \right| = 3.6 \cr
& {\text{Period: }}\frac{{2\pi }}{{\left| B \right|}} = \frac{{2\pi }}{{\left| {\pi /24} \right|}} = 48 \cr
& \cr
& {\text{amplitude: 3}}{\text{.6}} \cr
& {\text{period: 48}} \cr} $$
