Question 252405
Here is the first equation: 
{{{x^2- 9/x + 3}}}

Notice the variable in the denominator. We don't like these there and want to know where the function can't exist. Set the denominator not = 0 and solve. In this case, X not = 0. This gives us a vertical asymptote at x = 0.From here we create a table of values.
(X,Y): (-2,11.5), (-1, 13) (1,-.5) (2,2.5) (3,9) (4,16.5)
Notice the graph is a hyperbola.

Here is the second equation
{{{x^2 - x - 6/3}}}
Since there are no restrictions on X, we go right to a table of values.
(X,Y): (-3,10)(-2,4) (-1,0) (0,-2) (1,-2) (2,0) (3,4)
Notice this graph is a parabola.

Now I am not sure if you wanted to know where they crossed, if at all?

They would cross at {{{x = (-5 +- sqrt(61))/2}}}