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no [but open to debate]

((x-1)(x-2)(x+2))/(x-3) (x^2-3x+2)/(x-2)(x+2) Asymptote missing, graph it, there is no Asymptote because the (x-2)(x+2) can be factored out.

yes

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Q: Does every rational function have a vertical asymptote?
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Related questions

Does every rational function have more than one vertical asymp tote?

No.The equation x/(x^2 + 1) does not have a vertical asymptote.


Does every rational function have an asymptote?

No if the denominators cancel each other out there is no asymptote


Does every rational function has at least one horizontal asymptote?

Nope not all the rational functions have a horizontal asymptote


Is it true that the function has a vertical asymptote at every x value where its numerator is zero and you can make a table for each vertical asymptote to find out what happens to the function there?

Every function has a vertical asymptote at every values that don't belong to the domain of the function. After you find those values you have to study the value of the limit in that point and if the result is infinite, then you have an vertical asymptote in that value


Does every undefined value of fx lead to a vertical asymptote?

No. For example, in real numbers, the square root of negative numbers are not defined.


A function will always be a vertical line?

A function can never be a vertical line, because it then fails the definition of a function: every x value outputs only 1 y value. The vertical line test will determine if a relation is a function. If a vertical line intersects the graph of the function at more than one place, it is not a function.


Is the cardinality of an infinitely countable set the same as the rational numbers?

Yes. There is an injective function from rational numbers to positive rational numbers*. Every positive rational number can be written in lowest terms as a/b, so there is an injective function from positive rationals to pairs of positive integers. The function f(a,b) = a^2 + 2ab + b^2 + a + 3b maps maps every pair of positive integers (a,b) to a unique integer. So there is an injective function from rationals to integers. Since every integer is rational, the identity function is an injective function from integers to rationals. Then By the Cantor-Schroder-Bernstein theorem, there is a bijective function from rationals to integers, so the rationals are countably infinite. *This is left as an exercise for the reader.


Is there an instance when a linear equation is not a function?

Yes, a vertical line is linear, but it is not a function, because every point on the line has the same x value.


How is y 3x a function?

y=3x is a function because if you graph it, it passes the Vertical Line Test (draw a vertical line anywhere, a function will not touch the line twice or more). Also, it only has one y value for every x value.


Is every integer a rational number or is every rational number an integer?

Every integer is a rational number.


Why is y equals 3x a function?

y=3x is a function because if you graph it, it passes the Vertical Line Test (draw a vertical line anywhere, a function will not touch the line twice or more). Also, it only has one y value for every x value.


Why would you use the vertical line test?

To see if a graph is a function. There is only one x for every y!