Question

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

Answer 1

I would factor denominator and numerator (just in case there are common factors amongst the two) first

Answer 2

Yeah X+1 cancels. I worked out the problem I just want to see if I did it correctly.

Answer 3

ok so that means there is a hole at x=-1
what was the other factor in the denominator?

Answer 4

2x-3 and x+1 in the numerator

Answer 5

I got VA: -1 and HA: 3/2

Answer 6

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

Answer 7

(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

Answer 8

also leave your asymptotes in equation form because asymptotes are suppose to be described by equations not just numbers

Answer 9

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.

Answer 10

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
\[y=\frac{\color{red}{1}x^2+2x+1}{\color{red}2x^2-x-3}\]

Answer 11

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

Answer 12

Oh I got it, so 3/2 is the Va?

Answer 13

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

Answer 14

oops that x+2 is a type-o should be x+1 in the numerator

Answer 15

Alright I got it thanks!

Answer 16

np