Find a formula for a function that has vertical asymptotes x=3, x=5 and horizontal asymptote y=3.

Answer 1

In order to find a function that fits the criteria given, you need to know how to find the asymptotes in the first place.

Vertical Asymptotes: we set the denominator equal to 0 and solve for x

Horizontal Asymptotes: we compare the degree of the numerator and denominator; if the degrees are equal, the HA equals the ratio of the leading coefficients, which is the case for this question

For the vertical asymptotes, if x = 3 and x = 5, then we can write the factors as (x - 3) and (x - 5).

For the horizontal asymptotes, if y = 3, the degree of the numerator and denominator are equal so the ratio of the leading coefficients must be 3/1.

f(x) = (3x²) / [(x-3)(x-5)] which can be rewritten as f(x) = (3x²) / (x² - 8x + 15)

By double checking, in order for the denominator to equal 0, x must be 3 or 5 → VA: x=3, x=5

Because the numerator and denominator are the same degree, we take a ratio of leading coefficients, which is 3/1 → HA: y=3