Question

Find a function y = f(x) that has only one horizontal asymptote and

Answer 1

Using asymptote concepts, it is found that the function is:

$f(x) = \frac{10x^2}{x^2 + 2x - 3}$

• First we find the vertical asymptotes, which are the values for which the function is outside the domain.

• We consider a fraction, thus, deciding to place vertical asymptotes at $x = -1$ and at $x = 3$, the denominator is:

$(x + 1)(x - 3) = x^2 + 2x - 3$

• The horizontal asymptote is the limit of f(x) as x goes to infinity. We suppose it is 10, thus the numerator is $10x^2$, as it has to be the same degree of the denominator.

Which means that the function is:

$f(x) = \frac{10x^2}{x^2 + 2x - 3}$

The graph is sketched below.

A similar problem is given at brainly.com/question/17375447