Question 39325
{{{ f(x) = (x^2+3x-4)/(x^2+5x-6) }}}
{{{ f(x) = ((x+4)(x-1))/((x+6)(x-1)) }}}
{{{ f(x) = (x+4)/(x+6) }}}


so, we are to avoid x+6=0, which is when x=-6. This the vertical asymptote.


Now for the horizontal one:
{{{ y = (x+4)/(x+6) }}}
{{{ y(x+6) = x+4 }}}
{{{ xy+6y = x+4 }}}
{{{ xy-x = 4-6y }}}
{{{ x(y-1) = 4-6y }}}
{{{ x = (4-6y)/(y-1) }}}


so, we are to avoid y-1=0, which is when y=1. this is the horizontal asymptote.


So, plot the axes and draw the 2 dotted asymptotes. We need a couple of points to anchor the 2 halves of the plot.


So, when x=0, we have y=4/6 --> y=2/3
and when y=0, we have x=4/(-1) --> x=-4


Plot these 2 points and draw a curve in the quadrant joining these points and heading towards the asymptotes. Then draw a matching curve in the diagonally opposite quadrant of your graph.


If you want to, pick some random value of x, say x=-5 and find y. It should be a small-ish negative y-value, since the curve crosses the x-axis at x=-4.


jon.