Question 529116
Your work is already half done.
You are correct about the y-intercept.
The factoring
{{{y=(2(x-3))/((x+5)(x-2))}}}
tells you a lot about the function.
Look at when each factor in parenthesis is zero.
It says that for {{{x=3}}} {{{y=0}}}.
That's the x intercept, where the graph crosses the x-axis.
You also see from the factoring that the function changes sign each time one of those parentheses changes sign.
It is negative when all three parentheses are negative, for {{{x<-5}}}.
It becomes positive between -5 and 2.
It's negative again between 2 and 3,
and it's positive for {{{x>3}}}.
You also see that the function does not exist for
{{{x=-5}}} and for {{{x=2}}}
because they make the denominator zero. Those values of x are not part of the domain of the function, but other than that, all other real numbers are in the domain.
As x approaches -5, the factor {{{(x+5)}}} aproaches zero, while the other factors are close to finite numbers. Very near -5, the function could be approximated by
{{{(2(-5-3))/((x+5)(-5-2))=(16/7)(1/(x+5))}}}.
That means it goes down towards minus infinity on the left side of the vertical line x=5 and comes back down from plus infinity on the other side of -5. 
Something similar, but reversed happens around x=2.
Both, x=-5 and x=2 are vertical asymptotes.
Even more can be said about the behavior of the function.
For very large x, and for very large -x, y is very small. The function hugs the x- axis as you go to one or the other extremes of x. The x- axis is an asymptote on both sides (a horizontal one).
Maybe you could see that better if you wrote it as
{{{y=(2x-6)/((x+5)(x-2))=(2x-4)/((x+5)(x-2))-2/((x+5)(x-2))=2/(x+5)-2/((x+5)(x-2))}}}