Q: Create rational function that has the given lines as asympotes?

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deterministic

As given it is rational and a terminating decimal.

The given whole number of -2 is a rational number

Within a given range, the set of rational numbers is greater. Without a given range, both sets are infinite and a comparison is not very helpful.

There are an infinite number of rational numbers between any two given numbers.

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You cannot, necessarily. Given a graph of the tan function, you could not.

The function of creating a thumbnail image is that it help you to easily locate a given file.

deterministic

y=6/5x is a rational function not a line.

Add them together and divide by 2 will give one of the rational numbers between two given rational numbers.

function generator generates different types of functions to cro. And it can send analog signals to cro to create their graphical representation, and by it we can vary frequency and amplitude of the given wave function

function generator generates different types of functions to cro. And it can send analog signals to cro to create their graphical representation, and by it we can vary frequency and amplitude of the given wave function

function generator generates different types of functions to cro. And it can send analog signals to cro to create their graphical representation, and by it we can vary frequency and amplitude of the given wave function

As given it is rational and a terminating decimal.

There weren't any numbers given below.

The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.

The given whole number of -2 is a rational number