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anonymous

  • one year ago

f(x)= Solve for Vertical and horizontal Asymptote: x^2+2x+1 / 2x^2-x-3

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  1. freckles
    • one year ago
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    I would factor denominator and numerator (just in case there are common factors amongst the two) first

  2. anonymous
    • one year ago
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    Yeah X+1 cancels. I worked out the problem I just want to see if I did it correctly.

  3. freckles
    • one year ago
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    ok so that means there is a hole at x=-1 what was the other factor in the denominator?

  4. anonymous
    • one year ago
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    2x-3 and x+1 in the numerator

  5. anonymous
    • one year ago
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    I got VA: -1 and HA: 3/2

  6. freckles
    • one year ago
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    sounds good so you solved 2x-3=0 to get the vertical asymptote though you could probably just say 2x-3=0 but we know teachers like us to solve this for x now the horizontal asymptote is easier; you have the degrees are the same on top and bottom so you just fetch the coefficient of x^2 on top and bottom

  7. freckles
    • one year ago
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    (x+1)'s cancel so you do not have a vertical asymptote at x=-1 2x-3=0 will give you the vertical asymptote and I'm not sure how you got that horizontal asymptote

  8. freckles
    • one year ago
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    also leave your asymptotes in equation form because asymptotes are suppose to be described by equations not just numbers

  9. anonymous
    • one year ago
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    I used the 1st x+1 in the numerator for VA and 2x-3 in the denominator for HA setting them both equal to zero and the denominator is larger so I believe that would make the denominator be HA.

  10. freckles
    • one year ago
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    since the degree on top and bottom are both 2 then that means we just take the coefficients of the terms on top and bottom of the x^2 term to find the horizontal asymptote \[y=\frac{\color{red}{1}x^2+2x+1}{\color{red}2x^2-x-3}\]

  11. freckles
    • one year ago
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    the ha is y=1/2 as we said you do not have a va at x=-1 but you have a hole at x=-1 since the (x+1)'s canceled and you had no more (x+1)'s on bottom but you will have a va when 2x-3=0

  12. anonymous
    • one year ago
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    Oh I got it, so 3/2 is the Va?

  13. freckles
    • one year ago
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    \[f(x)=\frac{(x+1)(x+1)}{(x+1)(2x-3)}=\frac{x+2}{2x-3} , x \neq -1 \\ 2x-3=0 \text{ gives } x=\frac{3}{2} \text{ as the vertical asymptote } \\ \] don't just write 3/2 an asymptote is described by an equation not a number

  14. freckles
    • one year ago
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    oops that x+2 is a type-o should be x+1 in the numerator

  15. anonymous
    • one year ago
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    Alright I got it thanks!

  16. freckles
    • one year ago
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    np

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