im having a really difficult time finding out how to find the verticle asymptote and holes and horizontal asymptote. its really frustrating
for exmaple here are some questions:
thank you in advance
im having a really difficult time finding out how to find the verticle asymptote and holes and horizontal asymptote. its really frustrating
for exmaple here are some questions:
thank you in advance
let's start with something simple to see what you know ...
$\displaystyle y = \frac{(x-1)(x+2)(x+3)}{2(x+1)(x+2)(x-3)}$
where are the vertical asymptotes?
what is the horizontal asymptote?
where is the point discontinuity (hole) ?
horizontal asymptote at y = 1/2 ... it's the equation of a line
vertical asymptotes at x = -1 and x = 3 , "hole" at x = -2 ... note the common factor in the numerator and denominator.
OK ... using one of your problems
$\displaystyle g(x) = \frac{6x^2}{x^4 - 4x^2} = \frac{6x^2}{x^2(x^2 - 4)} = \frac{6x^2}{x^2(x-2)(x+2)}$
horizontal asymptote at y = ?
two vertical asymptotes ... where?
one "hole" ... where?
so horizontal asymptote @ y=6
verticle asymptotes at x=+2 and x=-2
the hole would be at.....X^2 would be the only common factor i see. but that would be sqrt0.... so there is no hole?
so after class today i got a good understanding of asymptotes and what not.
only thing that still doesnt make sense is holes. i understand that a hole occurs when you have common factors but what do i do to find the point and what does a hole mean?
A "hole" is really a point discontinuity ... the discontinuity occurs at the x-value that makes the common factor in the numerator and denominator equal to zero.
for example, the function $\displaystyle y = \frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{x-1}$ has a "hole" at $\displaystyle x = 1$. the graph of the function will look like the line $\displaystyle y = x+1$ except for the "hole" at $\displaystyle x = 1$ ...
note the graph made on my TI calculator ... see the "hole" ?
sorry, but no ...
there is a hole at x = -1 (the value of x that makes the factor common to the numerator and denominator 0), the y-value of the hole is 3/4
there is one vertical asymptote at x = 3