anonymous
  • anonymous
Describe the vertical asymptote(s) and hole(s) for the graph of y= (x+3)(x+4) / (x+4)(x+2). Answers: a) asymptote: x=3 and hole: x=-2 b) asymptote: x=-2 and hole: x= -4 c) asymptote: x=-2 and hole: x=4 d) asymptote: x=2 and hole: x=4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@nincompoop @Hero please help me
Hayleymeyer
  • Hayleymeyer
cross off what you Dont think is the answer
anonymous
  • anonymous
I WILL FAN + MEDAL PLEASE HELP !

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More answers

anonymous
  • anonymous
@dumbcow @hick4life
Nnesha
  • Nnesha
for \(\color{green}{\rm Vertical~ asy.}\) set the denominator equal to zero and then solve for the variable. for\(\color{green}{\rm Horizontal ~asy.}\) focus on highest degrees ~if the highest degree of the numerator is greater than the denominator then `No horizontal asy.` \[\color{reD}{\rm N}>\color{blue}{\rm D}\] example \[\large\rm \frac{ 7x^\color{ReD}{3} +1}{ 4x^\color{blue}{2}+3 }\] ~if the highest degree of the denominator is greater than the highest degree of the numerator then `y=0` would be horizontal asy. \[\rm \color{reD}{N}<\color{blue}{\rm D}\] example:\[\large\rm \frac{ 7x^\color{red}{2}+1 }{ 4x^\color{blue}{3}+3 }\] ~if both degrees are the same then divide the leading coefficient of the numerator by the leading coefficient of the denominator \[\rm \color{red}{N}=\color{blue}{D}\] \[\large\rm \frac{ 8x^\color{reD}{3}+1 }{ 4x^\color{blue}{3}+3 }\] \[\rm \frac{ 8x^3 }{ 4x^3 } =2\] horizontal asy. =2
hick4life
  • hick4life
ok lets see what we got here
anonymous
  • anonymous
i will medal and fan if you get me the answer @hick4life
Nnesha
  • Nnesha
\[\rm f(x)=\frac{ (x+3)(x+4)}{ (x+2)(x+4)}\] first simplify is there anything you can cacnel out?
Nnesha
  • Nnesha
cancel*
hick4life
  • hick4life
Do we have a graph line
dumbcow
  • dumbcow
@jordanaz26 , just getting the answer wont help you on the next question...
anonymous
  • anonymous
yeah we can cancel out (x+4) @Nnesha
Nnesha
  • Nnesha
yes right! so \[f(x)=\frac{ (x+3) }{ x+2}\]set the denominator equal to zero to find vertical asy
Nnesha
  • Nnesha
x+2=0 solve for x
hick4life
  • hick4life
Go ahead and cancel (4x+)
anonymous
  • anonymous
x=-2 @Nnesha
Nnesha
  • Nnesha
:=) looks good wait a sec plz
anonymous
  • anonymous
sure ! so we got the vertical asymptote which is x:-2, but it also asks for the hole(s) @Nnesha
hick4life
  • hick4life
Its going to be B.
Nnesha
  • Nnesha
yeah i'mm trying to find hole in my notebook i forgot how to find it
Nnesha
  • Nnesha
ahh okay what are the factors that in common to both numerator and denominator ?
anonymous
  • anonymous
of 3 and 2 ? @Nnesha
Nnesha
  • Nnesha
common factors would be hole \[\rm f(x)=\frac{ (x+3)\color{reD}{(x+4)}}{ (x+2)\color{red}{(x+4)}}\] in this question x+4 is common right so set it equal to zero x+4=0 solve for x that value would be hole
anonymous
  • anonymous
oh !!! so its x=-4 ! thank you so much ! fan and medal ! @Nnesha
Nnesha
  • Nnesha
my pleasure
Nnesha
  • Nnesha
and yes it's -4

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