Symmetry, periodicity and intersection points of a function

1. The function is symmetric with respect to the axis $x=0$: $$ f(-x)=(-x)^2-4=x^2-4=f(x)$$ The function is not periodic since it does not repeat itself.

Intersection points with the axes:

Si $x=0 \Rightarrow y=f(0)=-4 \Rightarrow (0,-4)$

Si $y=0 \Rightarrow 0=f(x)=x^2-4 \Rightarrow x^2=4 \Rightarrow x=\pm2 \Rightarrow (2,0), \ (-2,0)$

2. The function is symmetric with respect to the axis $x=0$ since $$\cos (x)=\cos(-x)$$ Also, the cosine is $2\pi$-periodic: $\cos(x+2\pi)=\cos(x)$.

Intersection points with the axes:

If $x=0 \Rightarrow y=f(0)=\cos(0)=1 \Rightarrow (0,1)$

If $y=0 \Rightarrow 0=f(x)=\cos(x) \Rightarrow x=\pi+\pi k \Rightarrow (\pi+\pi k,0)$ for $k\in\mathbb{Z}$

3. This function is antisymmetric with respect to the axis $x=0$ since $$ f(-x)=\dfrac{-2x}{(-x)^2-1}=\dfrac{-2x}{x^-1}=-f(x)$$ It does not present any type of period.

Intersection points with the axes:

If $x=0 \Rightarrow y=f(0)=0 \Rightarrow (0,0)$

If $y=0 \Rightarrow 0=f(x)=\dfrac{2x}{x^2-1} \Rightarrow 0=2x \Rightarrow x=0 \Rightarrow (0,0)$

4. Clearly antisymmetrical function in the axis $x=0$: $$f(-x)=-x=-f(x)$$ Intersection points with the axes:

If $x=0 \Rightarrow y=f(0)=0 \Rightarrow (0,0)$

If $y=0 \Rightarrow 0=f(x)=x \Rightarrow x=0 \Rightarrow (0,0)$