We have that the general equation of the hyperbola is the following:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]where 'a' and '-a' are the x-intercepts.
In this case, we have the following equation:
[tex]9x^2-16y^2=144[/tex]then, we have to divide both sides by 144 to get on the right side 1, thus, we have the following:
[tex]\begin{gathered} (9x^2-16y^2=144)(\frac{1}{144}) \\ \Rightarrow\frac{9}{144}x^2-\frac{16}{144}y^2=1 \\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1 \end{gathered}[/tex]now, notice that we have that:
[tex]a^2=16[/tex]then, the x-intercepts are:
[tex]\begin{gathered} a^2=16 \\ \Rightarrow a=\pm\sqrt[]{16}=\pm4 \\ a_1=4 \\ \text{and} \\ a_2=-4 \end{gathered}[/tex]therefore, the x-intercepts are 4 and -4.
Now that we know that the intercepts are on those points, we can see that the hyperbola has the following graph:
we can see that the hyperbola is not defined between -4 and 4, therefore, the domain of the hyperbola is:
[tex](-\infty,-4\rbrack\cup\lbrack4,\infty)[/tex]finally, to perform the symmetry test, we have to check first both axis simmetries by changing x to -x and y to -y:
[tex]\begin{gathered} \frac{x^2}{16}-\frac{(-y)^2}{9}=1 \\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1 \\ then \\ \frac{(-x)^2}{16}-\frac{y^2}{9}=1 \\ \Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1 \end{gathered}[/tex]since the hyperbola got symmetry about the x-axis and the y-axis, we have that the hyperbola got symmetry about the origin
