Answer
a)The intercept with $x$ (or $y$) axis is the point in which the line representing the graph of the function passes through $x$ (or $y$) axis. From the figure given in the problem we see that:
For $y=x^3-x$ the intercepts with $x$ axis are points $(-1,0)$ and $(1,0)$ and with $y$ axis the point $(0,0)$.
For $y=x^2+2$ the intercepts the graph does not cross the $x$ axis so there are no intercepts with it. The intercept with $y$ axis is at the point $(0,2)$.
b) The equation $$y=x^2+2$$ is symmetric with respect to $x$ axis. The equation $$y=x^3-x$$ is symmetric with respect to the origin.
c) The point of intersection $(2,6)$.
Work Step by Step
a)The intercept with $x$ (or $y$) axis is the point in which the line representing the graph of the function passes through $x$ (or $y$) axis. From the figure given in the problem we see that:
For $y=x^3-x$ the intercepts with $x$ axis are points $(-1,0)$ and $(1,0)$ and with $y$ axis the point $(0,0)$.
For $y=x^2+2$ the intercepts the graph does not cross the $x$ axis so there are no intercepts with it. The intercept with $y$ axis is at the point $(0,2)$.
b) The equation $$y=x^2+2$$ is symmetric with respect to $x$ axis since is we change $x$ to $-x$ the equation does not change:
$$y=(-x)^2+2=x^2+2$$
The equation $$y=x^3-x$$ is symmetric with respect to the origin since it does not change if we put both $-x$ instead of $x$ and $-y$ instead of $y$:
$$-y=(-x)^3-(-x)=-x^3+x.$$ Multiplying by $-1$ we obtain the original equation:
$$y=x^3-x$$
so it is symmetric with respect to the origin.
c) To determine the point of intersection we have to solve the system consisting of the two given equations (find $x$ and $y$ that satisfy both of them):
\begin{align}
y=x^2+2\\
y=x^3-x\\
\end{align}
Since both of the left sides are $y$ we can equate the right sides and calculate $x$:
$$x^3-x=x^2+2.$$
Rearranging terms
$$x^3-x^2-x-2=0.$$
This is the equation of third degree with whole number coefficients so we will have to guess one solution. Looking at the equation we see that it is satisfied by $x=2$:
$$2^3-2^2-2-2=8-4-2-2=0.$$
Now we can factor $$x^3-x^2-x-2=(x-2)(x^2+x+1).$$
The polynomial $x^2+x+1$ has no real zeros (since negative $x$ cannot balance out positive $x^2+1$) so the only solution for $x$ is $x=2$. Now returning this into, lets say, 1st equation we obtain
$$y=2^2+2=4+2=6.$$
Now we have for the point of intersection $(2,6)$.
