The general form of a trigonometric function is:
[tex]A\sin (B(x-C))+D[/tex]
Where B is the frequency of the function.
In our problem, A=1, C=D=0.
Then, as the value of B increases, so the frequency does. The answer to the second gap is 'increases'.
On the other hand, let P be the period and f the frequency. Those two quantities are related by the formula:
[tex]f=\frac{1}{P}[/tex]
Then, if the frequency increases, the period decreases. The answer to the first gap is 'decreases'.
Finally, if B is negative we have that:
[tex]\begin{gathered} B<0,A=-B,A>0 \\ \Rightarrow\tan (Bx)=\frac{\sin(Bx)}{\cos(Bx)}=\frac{\sin(-Ax)}{\cos(-Ax)}=-\frac{\sin(Ax)}{\cos(Ax)}=-\tan (-Bx) \end{gathered}[/tex]
Therefore, the function is reflected over the x-axis.
