asymptotes and holes of f(x) = [-3(x - 4)(x + 2)^2 (x + 1)(x

[Math wiz ya rite 09]

Write the equations of the asymptotes and give the coordinates of all holes for f(x).

f(x) = [-3(x - 4)(x + 2)^2 (x + 1)(x - 5)] / [(x + 3)^2 (x - 1)(x - 5)]

 

[skeeter]

factors that cancel out in the numerator and denominator give the location of "holes".
the x-value that makes the factor = 0 is the location.

factors in the denominator that do not cancel with any factor in the numerator give the location of vertical asymptotes. the x-value that makes these factors = 0 is the location (and equation) for vertical asymptotes.

for example ...

y = [x(x-1)(x+3)]/[x(x+1)(x-5)]

note that the lone "x" factor cancels ... there is a "hole" at x = 0

note that the factors (x+1) and (x-5) do not cancel with any factor in the numerator ... vertical asymptotes are x = -1 and x = 5

 

[Math wiz ya rite 09]

according to the above responce, i have concluded that there shall be;



vertical assymptote at x = -3

and holes at 5 and 1.

shouldn't holes be coordinate points though?


is there a horizontal assymptote?

did i do nething rite?

 

[soroban]

Re: asymptotes and holes

Hello, Math wiz ya rite 09!

Write the equations of the asymptotes and give the coordinates of all holes for \(\displaystyle f(x).\)

. . \(\displaystyle f(x) \:=\:\frac{-3(x\,-\,4)(x\,+\,2)^2(x\,+\,1)(x\,-\,5)}{(x\,+\,3)^2(x\,-\,1)(x\,-\,5)}\)

There are vertical asymptotes at: \(\displaystyle \,x\,=\,-3\) and \(\displaystyle x\,=\,1\)

There is a "hole" at \(\displaystyle x\,=\,5\)


To find the y-coordinate of the hole, reduce the fraction:

. . \(\displaystyle f(x)\;=\;\frac{-3(x\,-\,4)(x\,+\,2)^2(x\,+\,1)\sout{(x\,-\,5)}}{(x\,+\,3)^2(x\,-\,1)\sout{(x\,-\,5)}} \;=\;\frac{-3(x\,-\,1)(x\,+\,2)^2(x\,+\,1)}{(x\,+\,3)^2(x\,-\,1)}\)

Then: \(\displaystyle \:f(5)\;=\;\frac{-3(1)(7^2)(6)}{(8^2)(4)}\;=\;-\frac{441}{128}\)

Therefore, the hole is at: \(\displaystyle \,\left(5,\,-\frac{441}{128}\right)\)