anonymous
  • anonymous
f(x)= Solve for Vertical and horizontal Asymptote: x^2+2x+1 / 2x^2-x-3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
I would factor denominator and numerator (just in case there are common factors amongst the two) first
anonymous
  • anonymous
Yeah X+1 cancels. I worked out the problem I just want to see if I did it correctly.
freckles
  • freckles
ok so that means there is a hole at x=-1 what was the other factor in the denominator?

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anonymous
  • anonymous
2x-3 and x+1 in the numerator
anonymous
  • anonymous
I got VA: -1 and HA: 3/2
freckles
  • freckles
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
freckles
  • freckles
(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
freckles
  • freckles
also leave your asymptotes in equation form because asymptotes are suppose to be described by equations not just numbers
anonymous
  • anonymous
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.
freckles
  • freckles
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 \[y=\frac{\color{red}{1}x^2+2x+1}{\color{red}2x^2-x-3}\]
freckles
  • freckles
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
anonymous
  • anonymous
Oh I got it, so 3/2 is the Va?
freckles
  • freckles
\[f(x)=\frac{(x+1)(x+1)}{(x+1)(2x-3)}=\frac{x+2}{2x-3} , x \neq -1 \\ 2x-3=0 \text{ gives } x=\frac{3}{2} \text{ as the vertical asymptote } \\ \] don't just write 3/2 an asymptote is described by an equation not a number
freckles
  • freckles
oops that x+2 is a type-o should be x+1 in the numerator
anonymous
  • anonymous
Alright I got it thanks!
freckles
  • freckles
np

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