Given
[tex]f(x)=\tan x[/tex]
To fill the blanks.
Now,
The grpah of f(x)=tanx is given as,
Therefore, from the graph,
[tex]\tan x=\tan (x+\pi)=\tan (x+2\pi)[/tex]
Then,
[tex]\pi\text{ is the fundamental period of tanx.}[/tex]
Also,
For x=0, tan(0)=0.
Therefore, the y-intercept is (0,0).
The points for the function f(x)=tanx is,
[tex]\begin{gathered} x=\frac{\pi}{3},\text{ y=tan(}\frac{\pi}{3})=\sqrt[]{3} \\ x=\frac{\pi}{4},y=\tan \frac{\pi}{4}=1 \end{gathered}[/tex]
Since
[tex]\tan (-A)=-\tan A[/tex]
Then, the points of the graph y=tanx is,
[tex](-\frac{\pi}{3},-\sqrt[]{3}),(-\frac{\pi}{4},-1),(0,0),(\frac{\pi}{3},\sqrt[]{3}),(\frac{\pi}{4},1)[/tex]
Hence, the answer should be in the order,
i)
[tex]\pi[/tex]
ii)
[tex](0,0)[/tex]
iii)
[tex](-\frac{\pi}{3},-\sqrt[]{3}),(-\frac{\pi}{4},-1),(0,0),(\frac{\pi}{3},\sqrt[]{3}),(\frac{\pi}{4},1)[/tex]
