Help me solve this rational problem?

Given: #(x^2-16)/(x^2+x-20) #

There are several things to note.

#color(brown)("Consider the numerator:")#

#x^2-16 -> a^2-b^2=(a-b)(a+b)#

Thus we have #(x-4)(x+4)#

From this it MAY be the case that the question designer intended one of them to cancel out in the denominator.

#color(brown)("Consider the denominator:")#

As it is possible that there MAY be a cancelling out it is reasonable to explore #x=+-4# as a factor

Note that #4xx5=20 and 5-4=1# so lets look at

#x^2+x-20color(white)("ddd")->color(white)("ddd")(x+5)(x-4)#

#color(white)("dddddddddddd")->color(white)("ddd") x^2+5x-4x-20 larr" Works"#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Putting it all back together")#

#(x^2-16)/(x^2+x-20) -> (cancel((x-4))(x+4))/((x+5)cancel((x-4))) =(x+4)/(x+5) ->f(x)#

As the equation is undefined when the denominator becomes 0 we have a vertical asymptote at #x=-5#

# lim_(x->+oo) f(x)->k=+1 #

#lim_(x->-oo) f(x)->k==+1#

Thus there is a horizontal asymptote at #y=1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Any excluded value")#

Although the #(x-4)# in the denominator cancels 'away' it still forms part of the original expression. Thus it has to be taken into account. This is done by setting #(x-4)=0# and declaring #x=4# as the excluded value.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("The general behaviour of the plot")#

I will let you investigate this. You will need to shoe your logic in your solution.

We already know that there is a horizontal asymptote at #y=1#

Test out what happens very close to the vertical asymptote.

color(white)("d")

#"let "f(x)=(x^2-16)/(x^2+x-20)#

#"the numerator is a "color(blue)"difference of squares"#

#x^2-16=(x-4)(x+4)#

#"denominator"#

#"the factors of "-20" which sum to "+1#
#"are "+5" and "-4#

#x^2+x-20=(x+5)(x-4)#

#f(x)=(cancel((x-4))(x+4))/((x+5)cancel((x-4)))=(x+4)/(x+5)#

#"the removal of the factor "(x-4)" from the numerator"#

#" and denominator indicates a hole at"#

#x-4=0rArrx=4toy=(4+4)/(4+5)=8/9#

#"hole at "(4,8/9)#

#"the graph of "f(x)=(x+4)/(x+5)" is the same as"#

#(x^2-16)/(x^2+x-20)" but without the hole"#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "x+5=0rArrx=-5" is the asymptote"#

#"Horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by "x#

#f(x)=(x/x+4/x)/(x/x+5/x)=(1+4/x)/(1+5/x)#

#"as "xto+-oo,f(x)to(1+0)/(1+0)#

#y=1" is the asymptote"#

#"there is a discontinuity at "x=-5#

#"domain is "x in(-oo,-5)uu(-5,oo)#

#"there is a discontinuity at "y=1#

#"range is "y in(-oo,1)uu(1,oo)#

#"For Intercepts"#

#x=0rArry=4/5larrcolor(red)"y-intercept"#

#x+4=0rArrx=-4larrcolor(red)"x-intercept"#
graph{(x+4)/(x+5) [-10, 10, -5, 5]}