Question 816619
<pre>


f(x) = {{{(x^3-3x^2-10x)/(x^2+5x+6)}}}

Factor the numerator:     Factor the denominator

    x³-3x²-10x                x²+5x+6                    
    x(x²-3x-10)              (x+3)(x+2)
    x(x+3)(x+2)

f(x) = {{{(x(x-5)(x+2))/((x+3)(x+2))}}}

Since (x+3) is a factor of the denominator but not
the numerator, there is an asymptote where x+3=0,
or at x=-3, which is the equation of the vertical
asymptote, where there is a non-removable discontinuite.
 Since (x+2) is a factor of both denominator and numerator,
there is a removable discontinuity where x+2=0, at x=-2.

We may cancel the (x+2)'s as long as we also state that
x&#8800;2

f(x) = {{{x(x-5)/(x+3)}}}, where x&#8800;2

So we graph

y = {{{x(x-5)/(x+3)}}}, leaving a hole at x=2

There is a vertical asymptote at x=-3
Since the degree of the numberator is 1 more than the degree
of the denominator, there is no horizontal asymptote, but there
is an oblique (or slant) asymptote, which we find by long
division:

We have to multiply the numerator out and add +0 to divide:

y = {{{x²-5x+0)/(x+3)}}},

   <u>     x- 8</u>+{{{34/(x+3)}}}
x+3)x²-5x+ 0
    <u>x²+3x</u>
      -8x+ 0
      <u>-8x-24</u>
          34

Since the fraction {{{34/(x+3)}}} approaches 0 as x gets large,
the graph of f(x) must approach the line y=x-8, which is the
equation of the oblique (slant) asymptote.

We get the y-intercept by setting x = 0

y = {{{0(0-5)/(0+3)}}} = 0

So the y-intercept is (0,0)

We get the x-intercepts by setting y = 0 

0 = {{{x(x-5)/(x+3)}}}

0 = x(x-5)
    x=0;  x-5=0
            x=5

So the x-intercepts are (0,0) and (5,0)

We plot the asymptotes and the intercepts:


{{{drawing(200,800,-10,10,-50,30,graph(200,800,-10,10,-50,30),

circle(0,0,0.15),circle(0,0,0.13),circle(0,0,0.11),circle(0,0,0.09),circle(0,0,0.07),circle(0,0,0.05),circle(0,0,0.03),circle(0,0,0.01),
circle(5,0,0.15),circle(5,0,0.13),circle(5,0,0.11),circle(5,0,0.09),circle(5,0,0.07),circle(5,0,0.05),circle(5,0,0.03),circle(5,0,0.01),


green(line(-3,-70,-3,70),line(-27,-35,17,9)) )}}} 

Now we find any relative extrema points by
finding the derivative and setting it = 0

y = {{{(x(x-5))/(x+3)}}}
Multiply the top out:
y = {{{(x^2-5x)/(x+3)}}}
Use the quoptient formula for the derivative:
y' = {{{((x+3)(2x-5)-(x^2-5x)(1))/(x+3)^2}}}
y' = {{{(2x^2-5x+6x-15-x^2+5x)/(x+3)^2}}}
y' = {{{(x^2+6x-15)/(x+3)^2}}}
Setting that = 0 to find relative extrema:
{{{(x^2+6x-15)/(x+3)^2}}} = 0
x²+6x-15 = 0
Unfortunately that doesn't factor, so we must
use the quadratic formula:
{{{x = (-6 +- sqrt( 6^2-4*1*(-15) ))/(2*1) }}}
{{{x = (-6 +- sqrt(36+60))/2 }}}
{{{x = (-6 +- sqrt(96))/2 }}}
{{{x = (-6 +- sqrt(16*6))/2 }}}
{{{x = (-6 +- 4sqrt(6))/2 }}}
{{{x = (2(-3 +- 2sqrt(6)))/2 }}}
{{{x = (cross(2)(-3 +- 2sqrt(6)))/cross(2) }}}
x = -3 ± 2V6
Approximating:  x=-7.90 and x=1.90
Substuting those in y, we get approximately
                y=-20.8 and y=-1.20

Relative extrema candidates are approximately (-7.90,-20.8)
and (1.90,-1.20)

To find out whether they are relative maximums or minimums,
or any inflection points, we must find the second derivative:

y' = {{{(x^2+6x-15)/(x+3)^2}}}
Use the quotient formula:
y" = {{{((x+3)^2(2x+6)-(x^2+6x-15)2(x+3)(1))/(x+3)^4}}}
y" = {{{((x^2+6x+9)(2x+6)-(x^2+6x-15)(2x+6))/(x+3)^4}}}
y" = {{{((2x^3+18x^2+54x+54)-(2x^3+18x^2+6x-90))/(x+3)^4}}}  
y" = {{{(2x^3+18x^2+54x+54-2x^3-18x^2-6x+90)/(x+3)^4}}}
y" = {{{(48x+144)/(x+3)^4}}}
y" = {{{(48(x+3))/(x+3)^4}}}
y" = {{{48/(x+3)^3}}}

Substituting x=-7.90, y" comes out negative,
therefore the point (-7.90,-20.8) is a relative
maximum, since the curvature is downward

Substituting x=1.90, y" comes out positive,
therefore the point (1.90,-1.20) is a relative
minimum, since the curvature is upward
 
To find any inflection points we set y"=0

{{{48/(x+3)^3}}} = 0
48 = 0
A contradiction so there are no inflection points.

So we draw the graph:

{{{drawing(200,800,-10,10,-50,30,graph(200,800,-10,10,-50,30,(x^3-3x^2-10x)/(x^2+5x+6)),

circle(0,0,0.15),circle(0,0,0.13),circle(0,0,0.11),circle(0,0,0.09),circle(0,0,0.07),circle(0,0,0.05),circle(0,0,0.03),circle(0,0,0.01),
circle(5,0,0.15),circle(5,0,0.13),circle(5,0,0.11),circle(5,0,0.09),circle(5,0,0.07),circle(5,0,0.05),circle(5,0,0.03),circle(5,0,0.01),


green(line(-3,-70,-3,70),line(-27,-35,17,9)) )}}} 

What a terribly long and messy problem!

Edwin</pre>