Question 295155
To find the y-intercept, set x=0 and solve for y.
{{{f(0)=(0+1)/(0^2-3(0)-10)=1/-10=-1/10}}}
(0,-1/10)
To find the x-intercept, set y=0 and solve for x.
{{{0=(x+1)/(x^2-3x-10)}}}
{{{x+1=0}}}
{{{x=-1}}}
(-1,0)
To find vertical asymptotes, look for x where the denominator goes to zero.
{{{x^2-3x-10=0}}}
{{{(x+2)(x-5)=0}}}
Two solutions:
{{{x+2=0}}}
{{{x=-2}}}
.
.
.
{{{x-5=0}}}
{{{x=5}}}
To find horizontal asymptotes, divide all terms by the highest order of x.
{{{f(x)=(x/x^2+1/x^2)/(x^2/x^2-3x/x^2-10/x^2)}}}
{{{f(x)=(1/x+1/x^2)/(1-3/x-10/x^2)}}}
Now take the limit as x goes to {{{infinity}}}.

{{{lim( x->-infinity, f(x) )=(0+0)/(1-0-0)=0}}}

So there is a horizontal asymptote at y=0.
.
.
. 
{{{drawing( 300, 300, -10, 10, -2, 2,
green(line(-2,-10,-2,10)),
green(line(5,-10,5,10)),

 graph( 300, 300, -10, 10, -2, 2, (x+1)/(x^2-3x-10))) }}}