Question 1158027
Condition #1: A zero at x = 4 and at x = -2
How to satisfy it: include(x-4) and (x+2) as factors
 
Condition #2: A y-intercept at y = 16/3
How to satisfy it: after satisfying other conditions, add a constant factor K to satisfy {{{f(0)=16/3}}}
 
Condition #3: A vertical asymptote at x = 3
How to satisfy it: include (x-3) as a denominator
 
Tentative solution to satisfy all 3 conditions above: {{{f(x)=K(x-4)(x+2)/(x-3)}}}
{{{f(0)=K(-4)(2)/(-3)=K(8/3)}}}
to satisfy {{{f(0)=16/3}}} , we make {{{K(8/3)=16/3}}} --> {{{K=2}}} 
 
So far, the function we have is {{{f(x)=2(x-4)(x+2)/(x-3)}}} , and its graph looks like this: 
{{{drawing(300,300,-16,12,-20,20,
line(3,-20,3,20),locate(3.2,16,x=3),
graph(300,300,-16,12,-20,20,2(x-4)(x+2)/(x-3))
)}}}
 
Condition #4: A slant asymptote at {{{y = 2x + 2}}}
We need {{{lim(x->infinity,(f(x)-(2x+2)))=0}}}
{{{f(x)-(2x+2)=f(x)-2(x+1)}}}{{{"="}}}{{{2(x-4)(x+2)/(x-3)-2(x+1)}}}{{{"="}}}{{{
2((x-4)(x+2)-(x+1)(x-3))/(x-3)}}}{{{"="}}}{{{2(x^2-2x-8-(x^2-2x-3))/(x-3)
}}}{{{"="}}}{{{2(x^2-x^2-2x+2x-8+3)/(x-3)}}}{{{"="}}}{{{2(-5)/(x-3)}}}{{{"="}}}{{{-10/(x-3)}}}
{{{lim(x->infinity,(f(x)-(2x+2)))=lim(x->infinity,(-10/(x-3)))=0}}}