Create a (simplified) rational function that has the following asymptotes.(show work) y=0, x=-3, x=0, and x=2
Hmmmm. f(x)= e^-x * 1/[(x+3)(x)(x-2)]
I think rational functions are ratios of polynomials, so forget the e^-x and just use
1/[(x+3)(x)(x-2)]
or, the numerator could be any polynomial of degree less than 3. It will still approach y=0 as x gets large.
To create a rational function with the given asymptotes, you need to consider the properties of a rational function and how they relate to the asymptotes.
A rational function consists of a polynomial in the numerator and a polynomial in the denominator. The degree of the polynomials determines the behavior of the function near the asymptotes.
Let's break down each asymptote:
1. Vertical asymptote at x = -3:
For a vertical asymptote at x = -3, the denominator should have a factor of (x + 3).
So, our function can be written as 1/(x + 3).
2. Vertical asymptote at x = 0:
For a vertical asymptote at x = 0, the denominator should have a factor of x.
So, our function can be written as 1/(x * (x + 3)).
3. Vertical asymptote at x = 2:
For a vertical asymptote at x = 2, the denominator should have a factor of (x - 2).
So, our function can be written as 1/((x - 2) * x * (x + 3)).
4. Horizontal asymptote at y = 0:
Since the highest power of x in the denominator (degree 3) is greater than the highest power of x in the numerator (degree 0), the horizontal asymptote occurs at y = 0.
Putting it all together, a simplified rational function with the given asymptotes is:
f(x) = 1/((x - 2) * x * (x + 3))
Please note that this is a simplified version of a rational function that satisfies the given asymptotes. The actual form of the function can vary depending on other factors, such as the location of zeros and the behavior at the local extrema.