Question 740836
1) {{{y= (x+3)/((x+2)(x+3))}}}

{{{y= cross((x+3))1/((x+2)cross((x+3)))}}}

{{{y= 1/(x+2)}}}



{{{Vertical}}} asymptotes will occur for {{{x-values}}} which make the denominator {{{zero}}}. For this equation the denominator will be zero if {{{x = -2}}} . So the vertical lines, {{{x =- 2}}} will be {{{vertical}}}{{{ asymptote}}} for this equation.

For this equation, when {{{x-values}}} become very large, the fraction approaches {{{zero}}} in value. 

so, {{{y=0}}} is {{{horizontal}}} asymptote

{{{ drawing(600, 600, -10, 10, -10, 20,  blue(line(-2,20,-2,-10)), graph( 600, 600, -10, 10, -10, 20, (x+3)/((x+2)(x+3)),0)) }}}




2.

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

{{{y=(x+5)/(x^2-5x+6)}}}
  

"Holes" would occur when the numerator and denominator have a common factor which is an expression which could be zero. 
Since this fraction has {{{no}}} such {{{common}}}{{{ factors}}}, there will be no "holes".

{{{Vertical}}} asymptotes will occur for {{{x-values}}} which make the denominator {{{zero}}}. For this equation the denominator will be zero if {{{x = 2}}} or {{{x = 3}}}. So the vertical lines, {{{x = 2}}} and {{{x = 3}}} will be {{{vertical}}}{{{ asymptotes}}} for this equation.

{{{Horizontal}}} asymptotes will occur if {{{y}}} approaches some constant value when {{{x-values}}} become {{{very}}}{{{ large}}} (positive or negative).
For this equation, when {{{x-values}}} become very large, the denominator of the fraction becomes very large. 
This makes the fraction very, very {{{small}}}. In fact the fraction approaches {{{zero}}} in value. 
So as {{{x-values}}} become very large, the fraction becomes negligible and the {{{y-value}}} approaches {{{x=0}}}.
 


{{{ drawing(600, 600, -10, 10, -10, 20,  blue(line(2,20,2,-10)),blue(line(3,20,3,-10)), graph( 600, 600, -10, 10, -10, 20, (x+5)/(x^2-5x+6),0)) }}}