# Create a function with the following properties: a. It has a horizontal asymptote at y= 2. b. It...

## Question:

Create a function with the following properties:

a. It has a horizontal asymptote at y= 2.

b. It has a discontinuity at x= 2 which is not a vertical asymptote.

c. It has no other discontinuities or asymptotes.

## Constructing a Rational Function

Since we know how horizontal and vertical asymptotes are defined, we can construct a rational function that has specific asymptotes. If it is supposed to be undefined at a value, we need to ensure that the denominator is zero at that number. However, whether it has a vertical asymptote or is just discontinuous at that point depends on whether or not the numerator is also zero at that point. If it isn't, then that point is a vertical asymptote, and if it is, then it is just discontinuous there.

To fulfill these conditions, we can construct a rational function. First, we need to examine what these conditions mean for that type of a function.

a. When the horizontal asymptote of a rational function is a nonzero constant, that means that the degree of the numerator is equal to the degree of the denominator. The value of this constant is the ratio of the leading coefficients. Thus, the numerator and denominator of this rational function must have the same degree, and the ratio of their leading coefficients must be 2.

b. Since the discontinuity at 2 isn't a vertical asymptote, the numerator and denominator must both equal zero when x equals 2. Thus, the term {eq}x - 2 {/eq} must be present in both.

c. Since there aren't any other discontinuities or asymptotes, the denominator cannot be zero anywhere else. The easiest way to ensure this is to not add any other expressions besides what we've already defined needs to be there.

The most simple rational function that satisfies these requirements is:

{eq}\begin{align*} y &=\frac{2(x-2)}{x-2}\\ &= \frac{2x-4}{x-2} \end{align*} {/eq}

The degree of the numerator and denominator are the same, and the ratio of the leading coefficients is 2, so the horizontal asymptote of {eq}y = 2 {/eq} exists. The numerator and denominator are both zero at {eq}x = 2 {/eq}, so this is a discontinuity but not a vertical asymptote. The denominator is not zero anywhere else, so there are no other discontinuities or asymptotes. Therefore, this function meets our requirements. 