Question

find vertical asymptote of f(x)= ((((x)^2)+(x)-(6)))/(((x)^2)-(x)-(2)))

Answer 1

@satellite73 please help!

Answer 2

can't read it, but whatever it is, set the denominator equal to zero and solve for \(x\)

Answer 3

looks like the denominator is \(x^2-x-2\) put
\[x^2-x-2=0\\
(x-2)(x+1)=0\] etc

Answer 4

\[f(x) =\frac{ x^2 +x-6}{x^2-x-2 ? }\]

Answer 5

yeah, ok, what i thought it might be
i wrote most of the solution above
you good from there?

Answer 6

no.. here's what i got…
when x=2 i get \[2^2+2-6 = 0\]
when x=-1 i get \[(-1)^2 + (-1)-6 = -6 \neq 0\]

correct?

Answer 7

lets go slow

Answer 8

for a rational function, the vertical asymptote is at the zeros of the denominator
you cannot divide by zero

Answer 9

the denominator of your rational function is \(x^2-x-2\) right?

Answer 10

right

Answer 11

therefore this question translates as "set \(x^2-x-2=0\) and solve for \(x\)"

Answer 12

luckily for you, this one factors
you can solve it by factoring as
\[(x-2)(x+1)=0\] and that means \(x=2\) or \(x=-1\)
those are both the zeros of the denominator, and also your vertical asymptotes

Answer 13

okay so i don't have to take the limits?

Answer 14

nope
it just asks for vertical asymptotes
the limits are either infinity or minus infinity, which are not numbers

Answer 15

ok so i also have to get the H.A. for the same problem… can you tell me if i did it correctly?

Answer 16

sure

Answer 17

\[f(x) =\frac{ x^2 +x-6}{x^2-x-2 }\] just to refresh my memory

Answer 18

\[\lim_{x \rightarrow \infty} \frac{ x^2+x-6 }{x^2-x-2} \frac{ \infty }{ \infty } form\]

Answer 19

tmi
degree of the numerator is the same as the degree of the denominator (they are both degree 2) so it is the ratio of the leading coefficients, which in this case is \(\frac{1}{1}=1\)

Answer 20

so i took the highest power in denominator and after my workings by taking out limit i got

Answer 21

i havent done that in class yet.. i was told i had to use highest power

Answer 22

but i did get the answer to be 1 so perfect

Answer 23

right
the highest power of both is 2
since they are the same, take the ratio of the leading coeficients
you can do it in your head
for example the horizontal asymptote of
\[\frac{4x^3+2x^2+x-1}{5x^3+20x-3}\] is \(y=\frac{4}{5}\)

Answer 24

ohhhhh. ok.. makes sense now. sorry. so if i did out the limits for my V.A.s i should get infinity?

Answer 25

oh yes
either \(\infty\) or \(-\infty\)

Answer 26

ok so i had to find va and ha for \[f(x) = \frac{ x+1 }{x^2-1}\] and i got va… x=1 and ha…. y=0….. is this correct?