Find an equation of a rational function f that satisfies the given conditions. vertical asymptotes: x = −3, x = 5 horizontal asymptote: y = 2 x-intercept: −5, 3; hole at x = 0

Answer 1

The vertical asymptotes come from the zeroes of the denominator.

x = -3

x + 3 = 0

x = 5

x - 5 = 0

(x + 3)(x - 5) = 0

For the horizontal asymptote to be 2, the leading degree of the numerator and denominator have to be the same and the numerator/denominator coefficient has to equal 2, like 2/1 or 4/2, etc. Pair that with a hole at x = 0 (where x - 0 exists in both the numerator and the denominator), and since we don't normally write x - 0, we have 2x/x.

2x/(x(x + 3)(x - 5))

Remember that y = 0 along the x-axis, so our x-intercepts are where y = 0. We can also call them the zeroes of the numerator.

x = -5

x + 5 = 0

x = 3

x - 3 = 0

(x + 5)(x - 3) = 0

Throw that allllll together and...

(2x(x + 5)(x - 3))/(x(x + 3)(x - 5))

If you're needing to multiply everything...

(2x³ + 4x² - 30x)/(x³ - 2x² - 30x) = 0