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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    @nincompoop @Hero please help me

  2. Hayleymeyer
    • one year ago
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    cross off what you Dont think is the answer

  3. anonymous
    • one year ago
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    I WILL FAN + MEDAL PLEASE HELP !

  4. anonymous
    • one year ago
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    @dumbcow @hick4life

  5. Nnesha
    • one year ago
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    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

  6. hick4life
    • one year ago
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    ok lets see what we got here

  7. anonymous
    • one year ago
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    i will medal and fan if you get me the answer @hick4life

  8. Nnesha
    • one year ago
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    \[\rm f(x)=\frac{ (x+3)(x+4)}{ (x+2)(x+4)}\] first simplify is there anything you can cacnel out?

  9. Nnesha
    • one year ago
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    cancel*

  10. hick4life
    • one year ago
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    Do we have a graph line

  11. dumbcow
    • one year ago
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    @jordanaz26 , just getting the answer wont help you on the next question...

  12. anonymous
    • one year ago
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    yeah we can cancel out (x+4) @Nnesha

  13. Nnesha
    • one year ago
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    yes right! so \[f(x)=\frac{ (x+3) }{ x+2}\]set the denominator equal to zero to find vertical asy

  14. Nnesha
    • one year ago
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    x+2=0 solve for x

  15. hick4life
    • one year ago
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    Go ahead and cancel (4x+)

  16. anonymous
    • one year ago
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    x=-2 @Nnesha

  17. Nnesha
    • one year ago
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    :=) looks good wait a sec plz

  18. anonymous
    • one year ago
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    sure ! so we got the vertical asymptote which is x:-2, but it also asks for the hole(s) @Nnesha

  19. hick4life
    • one year ago
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    Its going to be B.

  20. Nnesha
    • one year ago
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    yeah i'mm trying to find hole in my notebook i forgot how to find it

  21. Nnesha
    • one year ago
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    ahh okay what are the factors that in common to both numerator and denominator ?

  22. anonymous
    • one year ago
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    of 3 and 2 ? @Nnesha

  23. Nnesha
    • one year ago
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    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

  24. anonymous
    • one year ago
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    oh !!! so its x=-4 ! thank you so much ! fan and medal ! @Nnesha

  25. Nnesha
    • one year ago
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    my pleasure

  26. Nnesha
    • one year ago
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    and yes it's -4

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