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anonymous
 one year ago
f(x)= Solve for Vertical and horizontal Asymptote:
x^2+2x+1 / 2x^2x3
anonymous
 one year ago
f(x)= Solve for Vertical and horizontal Asymptote: x^2+2x+1 / 2x^2x3

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1I would factor denominator and numerator (just in case there are common factors amongst the two) first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah X+1 cancels. I worked out the problem I just want to see if I did it correctly.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok so that means there is a hole at x=1 what was the other factor in the denominator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02x3 and x+1 in the numerator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got VA: 1 and HA: 3/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sounds good so you solved 2x3=0 to get the vertical asymptote though you could probably just say 2x3=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
 one year ago
Best ResponseYou've already chosen the best response.1(x+1)'s cancel so you do not have a vertical asymptote at x=1 2x3=0 will give you the vertical asymptote and I'm not sure how you got that horizontal asymptote

freckles
 one year ago
Best ResponseYou've already chosen the best response.1also leave your asymptotes in equation form because asymptotes are suppose to be described by equations not just numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I used the 1st x+1 in the numerator for VA and 2x3 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
 one year ago
Best ResponseYou've already chosen the best response.1since 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 \[y=\frac{\color{red}{1}x^2+2x+1}{\color{red}2x^2x3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1the 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 2x3=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I got it, so 3/2 is the Va?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\frac{(x+1)(x+1)}{(x+1)(2x3)}=\frac{x+2}{2x3} , x \neq 1 \\ 2x3=0 \text{ gives } x=\frac{3}{2} \text{ as the vertical asymptote } \\ \] don't just write 3/2 an asymptote is described by an equation not a number

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops that x+2 is a typeo should be x+1 in the numerator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright I got it thanks!
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