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

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

Mathematics
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I would factor denominator and numerator (just in case there are common factors amongst the two) first
Yeah X+1 cancels. I worked out the problem I just want to see if I did it correctly.
ok so that means there is a hole at x=-1 what was the other factor in the denominator?

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Other answers:

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

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