Question 1157922
{{{F(x)=(x^2-7x-8)/(x+4)}}}

{{{F(x)=(x^2+x-8x-8)/(x+4)}}}

{{{F(x)=((x^2+x)-(8x+8))/(x+4)}}}

{{{F(x)=(x(x+1)-8(x+1))/(x+4)}}}

{{{F(x)=((x-8)(x+1))/(x+4)}}}


x-intercepts: set {{{F(x)=0}}}

{{{((x-8)(x+1))/(x+4)}}}

=>{{{x=8}}} and {{{x=-1}}}

x-intercepts at: ({{{8}}},{{{0}}}), ({{{-1}}},{{{0}}})


y-intercepts: set {{{F(x)=y}}} and {{{x=0}}}

{{{y=((0-8)(0+1))/(0+4)}}}

{{{y=(-8*1)/4}}}

{{{y=-2}}}

y-intercept at: ({{{0}}},{{{-2}}})


critical points: values of {{{x}}} for which is {{{f}}}’{{{(x)=0}}}

{{{F}}}'{{{(x) = (x^2 + 8 x - 20)/(x + 4)^2}}}

=> {{{ (x^2 + 8 x - 20)/(x + 4)^2=0}}}...only if numerator equal to zero
=> {{{(x + 10) (x - 2)=0}}}

 =>{{{x=-10}}}, {{{x=2}}}


vertical asymptote: {{{x=-4}}} (makes denominator zero)

Calculate the limits:

{{{lim(x->infinity,(x^2-7x-8)/(x+4))=infinity }}}

{{{lim(x->-infinity,(x^2-7x-8)/(x+4))=-infinity }}}


Thus, there are {{{no}}}{{{ horizontal}}}{{{ asymptotes}}}.


Do polynomial long division:

..........({{{x-11}}}
 {{{x+4}}}|{{{x^2-7x-8}}}
..........{{{x^2+4x}}}......subtract
............... {{{-11x-8}}}
............... {{{-11x-44}}}......subtract
.............................{{{ 36}}}

The rational term approaches {{{0}}} as the variable approaches{{{ infinity}}}.
Thus, the slant asymptote is: {{{x-11}}}


to find max and min, plug in critical points{{{x=-10}}}, {{{x=2}}}

{{{(x^2 - 7 x - 8)/(x + 4)= ((-10)^2 - 7 (-10) - 8)/(-10 + 4)=162/-6= -27}}} at {{{x = -10}}}->max

{{{(x^2 - 7 x - 8)/(x + 4)= (2^2 - 7 (2) - 8)/(2 + 4)=-18/6= -3}}} at{{{ x = 2}}}->minimum


 
{{{drawing ( 600, 600, -50, 50, -50, 50,
green(line(-4,50,-4,-50)),
graph( 600, 600, -50, 50, -50, 50, x-11, (x^2-7x-8)/(x+4))) }}}