How do you graph #f(x)=(x^3-2x^2-3x)/(4x^2+8x)# using holes, vertical and horizontal asymptotes, x and y intercepts?

We can factor both the top and the bottom to get asymptotes, some intercepts, and holes.

#x^3 - 2x^2 - 3x = x(x^2 - 2x-3) = x(x-3)(x+1)#
#4x^2 + 8x = 4x(x+2) #

We see that they both have a zero in common (x=0), which means that #x=0# is the one hole. If we consider the equation without the x term, we see that the function WOULD go through #(0, -3 / 8)#, which is where the hole (and technically the y-intercept) is.

Since the bottom also goes to zero at #x=-2#, that's a vertical asymptote. Since the top goes to zero at #x = 3, -1# those are zeros.

Last thing we need to observe: the top and bottom are different orders. Since the top is a higher order, in the limit, the equation will look like #y=x/4#.

Now we have everything except exact signs.
At #xrightarrow -infty#, the value is clearly negative so
#y < 0 forall x < -2#.

It then goes toward negative infinity and jumps to positive infinity after -2, until the next zero, i.e.
#y > 0 forall -2 < x < -1#.

At x=-1, it hits zero and switches sign, i.e.
#y < 0 forall -1 < x < 0 or 0 < x < 3#.

After x=3, the sign doesn't switch and the function quickly approaches that line, i.e.
# y > 0 forall x>3# and #y rightarrow x/4 #.

From all of that analysis, you should be able to sketch a plot similar to this: graph{x(x-3)(x+1)/(4x(x+2)) [-13.86, 13.86, -6.93, 6.93]}