The given equation of the ellipse is:

To find the x intercept, let y = 0
![\begin{gathered} x^2+4(0)^2\text{ = 4} \\ x\text{ = }\sqrt[]{4} \\ x\text{ = }\pm2 \\ x-\text{intercept = (}\pm2,\text{ 0)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/zvfxhhmfrfm08ocw4iosmlh3oz434mo7zk.png)
For the y-intercept, let x = 0
![\begin{gathered} 0^2+4y^2=4 \\ y^2\text{ = }(4)/(4) \\ y\text{ = }\sqrt[]{1} \\ y\text{ = }\pm1 \\ y-\text{intercept = (0, }\pm1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/d7vrvntm01yln0v2k9xrvcdjh1vmbrt8in.png)
For an ellipse of the form:

The given equation is:

![\begin{gathered} a^2\text{ = 4} \\ a\text{ = }\sqrt[]{4} \\ a\text{ = -2, 2} \\ b^2\text{ = 1} \\ b\text{ = }\sqrt[]{1} \\ b\text{ = -1, 1} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fz4dq4talukt6ui8hdpypqk9gsdeyx868q.png)
The x-domain = (-2, 2)
The y-domain = (-1, 1)
Symmetry test:
For the equation to be symmetriacal to the x-axis, y → -y

Because (1) matches with (2), the equation is symmetrical to the x-axis
For the equation to be symmetrical to the y-axis, x → -x

Since (1) matches with (2), the equation is symmetrical to the y-axis
For the equation to be symmetrical to the origin, x→-x, y→-y

The equation is symmetrical to the origin
Therefore, according to the symmetry test, the equation given is symmetrical to the x-axis, y-axis, and the origin