Question

How could you use the knowledge from this lesson to create a rational function that has a horizontal asymptote at y=2, a horizontal asymptote at x=-1, and a jump discontinuity at x=3?

Answer 1

Here is one way to create a rational function with the given asymptotes and discontinuity:

To have a horizontal asymptote at y=2, the numerator must have a degree less than the denominator. For example:

f(x) = x / (x^2 + bx + c)

To have a horizontal asymptote at x=-1, the denominator must be able to be factored into (x+1)(x+d). Let's set d=1.

f(x) = x / (x+1)(x+1)

To create a jump discontinuity at x=3, we can make the numerator 0 at x=3. Adding (x-3) to the numerator gives:

f(x) = (x-3)(x) / (x+1)(x+1)

Simplifying this gives:

f(x) = (x-3) / (x+1)

Which meets the criteria:

Horizontal asymptote at y=2

Horizontal asymptote at x=-1

Jump discontinuity at x=3

So in summary, one rational function meeting the given criteria is:

f(x) = (x-3) / (x+1)

Explanation:

Have a great day!