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

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No, this is not a function. The graph would have a vertical line at x=-14. Since there are more than one y value for every given x value, the equation does not represent a function. The slope of the equation also does not exist.

If for every point on the horizontal axis, the graph has one and only one point corresponding to the vertical axis; then it represents a function. Functions can not have discontinuities along the horizontal axis. Functions must return unambiguous deterministic results.

That's true. If a function is continuous, it's (Riemman) integrable, but the converse is not true.

No. A function is a "graph" the survives the "vertical line test". Namely, it is for every x in its domain, there can be one and only one f(x) in its co-domain. An ellipse clearly fails it at everywhere except it's two vertex. But an ellipse can be thought as two separate functions. A standard ellipse relation, x^2 / a + (y)^2 / b = 1, can be thought as two separate real functions of y1 and y2. where y1 = -y2 exactly.

For every vertical foot of pipe, take that and multiply it by .434.This equation will get within 1/2 -1/4 psi.

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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.

Nope not all the rational functions have a horizontal asymptote

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

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.

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

A graph is a function if every input (x-value) corresponds to only one output (y-value). One way to check for this is to perform the vertical line test: if a vertical line intersects the graph at more than one point, the graph 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.

Every integer is a rational number.

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.

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.

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

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