Question 669204
{{{F(x)=5-1/(x+4)^2}}} is related to the functions {{{f(x)=1/x}}} and {{{g(x)=1/x^2}}}.
You know that the function {{{f(x)=1/x}}} has a graph like this:
{{{graph(300,300,-5,5,-5,5,1/x)}}}.
The x-axis ({{{y=0}}}) is a horizontal asymptote that the function {{{f(x)=1/x}}} approaches as {{{x}}} and {{{-x}}} become larger and larger,
and you know that the function
does not exist for {{{x=0}}} because you cannot divide by zero.
Instead, you have a vertical asymptote at {{{x=0}}},
and there is no limit to the values {{{f(x)}}} can take.
It is never zero, but can take any other value.
 
Something similar happens with the function  {{{g(x)=1/x^2}}},
which does not exist for {{{x=0}}} and graphs as {{{graph(300,300,-5,5,-5,5,1/x^2)}}}
The only difference is that {{{g(x)=1/x^2}}} is never negative.
The domain is all real numbers except zero; the range is all {{{y>0}}}, and the asymptotes are the same.
 
The function {{{h(x)=1/(x+4)^2}}} is like {{{g(x)=1/x^2}}},
but everything is shifted 4 units to the left.
What happened to {{{g(x)=1/x^2}}} for {{{x=0}}}, {{{x=1}}}, and {{{x}}}
now happens for {{{h(x)=1/(x+4)^2}}} at {{{x=-4}}}, {{{x=-3}}}, and {{{x-4}}}.
The graph for {{{h(x)=1/(x+4)^2}}} is {{{graph(300,300,-9,1,-5,5,1/(x+4)^2)}}}.
The vertical asymptote is {{{x=-4),
and the function does not exist for {{{x=-4}}}.
(All other real numbers are in the domain of {{{f(x)}}}).
 
Adding a minus sign in front just flips the graph.
Everything becomes the reflection under the x-axis:
{{{graph(300,300,-9,1,-5,5,-1/(x+4)^2)}}}
 
Adding {{{5}}} to that, just moves the graph (and asymptotes) up by 5 units.
{{{graph(300,300,-9,1,-3,7,5-1/(x+4)^2,5)}}}
The domain is all the real numbers except {{{x=-4}}}.
The vertical asymptote is {{{x=-4}}}.
The range is all {{{y<5}}}.
The horizontal asymptote (shown in green) is {{{y=5}}}.