Question 1177854

{{{f(x)= (x-5)/(4-x^2)}}}

x - intercept(s):

set {{{f(x)=0 }}}
{{{(x-5)/(4-x^2)=0}}}..........since denominator cannot be zero, use numerator only

{{{(x-5)=0}}} =>{{{x=5}}}
x - intercept is at ({{{5}}},{{{0}}})


y - intercept:
set {{{x=0 }}}
{{{f(x)= (0-5)/(4-0^2)}}}
{{{f(x)= -5/4}}}
y - intercept is at ({{{0}}},{{{-5/4}}})


Vertical Asymptote(s):

{{{f(x)=(x-5)/(4-x^2)}}}.......set denominator equal to zero

{{{(4-x^2)=0}}}=>{{{4=x^2}}}=>{{{ x=2}}} and {{{x=-2}}}


Horizontal Asymptote(s):

{{{f(x)= (x-5)/(4-x^2)}}}

Degree of numerator is less than degree of denominator: horizontal asymptote at {{{f(x) = 0}}}.


Domain:
{ {{{x}}} element {{{R}}} : {{{x<>-2 }}}and {{{x<>2}}} }
or
{{{-2<x<2}}}


range:
{{{f(x)=y}}} 
{{{y= (x-5)/(4-x^2)}}}
Rewrite the above equation for x in standard form and solve using quadratic formulae
{{{y(4-x^2)= (x-5)}}}
{{{4y-yx^2= x-5}}}
{{{0= yx^2+x-4y-5}}}
{{{0= yx^2+x-(4y+5)}}}-> a=y, b=1, c=-(4y+5)
Find the discriminant to the above equation
{{{b^2-4ac=1-4*y(-(4y+5))}}}
.............={{{1+16*y^2+20y}}} 
.............={{{16*y^2+20y +1}}}

Using the quadratic formulas, the above equation gives the solutions

{{{x[1,2]=(1+-sqrt(16*y^2+20y +1))/2y}}}

The solutions {{{x[1,2]}}} are real if {{{16*y^2+20y +1  >=  0}}} and {{{y }}}≠{{{0}}}. 

Hence we need to solve the inequality
{{{16*y^2+20y +1 >= 0}}} and solution will be
{{{y>= (sqrt(21) - 5)/8}}}
{{{y<= (-5 - sqrt(21))/8}}}


End-Behavior:

function end-behavior of {{{(x-5)/(4-x^2)}}}:

as {{{x->infinity}}}, {{{f(x) ->0}}}, and
as {{{x->-infinity}}},{{{f(x)->0}}}


Intervals of Increasing/Decreasing:
If {{{f}}}' {{{(x )>0}}} then {{{f (x )}}} is increasing 
If {{{f}}}' {{{(x )<0}}} then {{{f (x ) }}}is decreasing 

{{{f(x)=(x-5)/(4-x^2)}}}...derivate

{{{f}}}'{{{(x) = (x^2 - 10 x + 4)/(x^2 - 4)^2}}}

Increasing {{{-infinity < x <-2}}}, and {{{-2< x <-sqrt(21)+5}}},
Decreasing:{{{-sqrt(21)+5 < x < 2}}}


Intervals of Positive/Negative:

positive when {{{ -infinity < x< -2 }}} U  {{{2< x < infinity }}}
negative when  {{{-2< x< 2}}}

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