Help!!!!!!?

Question

Construct one function (call it f) with all of the following properties:

i) f has exactly one vertical asymptote located at x=(3/2)
ii) f has an oblique asymptote at y=18x+28
iii) f has exactly two roots, both of which are non-real
iv) f has exactly two point discontinuities, one at x=(4/3) and one at x=-15/2)
v) f has a y-intercept at (0,-1)

Answer 1

1) Vertical asymptote: denominator has a factor of (x - 3/2)
K/(x - 3/2)

2) Oblique asymptote: Put appropriate powers of x in the numerator:
K(18x^2 + 28x + C)/(x - 3/2)

4) Multiply in factors of (x - 4/3)/(x - 4/3) and (x + 15/2)/(x + 15/2)
K((18x^2 + 28x + C)(x - 4/3)(x + 15/2))/((x - 3/2)(x - 4/3)(x + 15/2))

3) Quadratic in numerator needs C chosen so discriminent is negative:
28^2 - 4*18*C < 0, 784 - 72C < 0, 784 < 72C, so let C = 100 (plenty bif enough)
K((18x^2 + 28x + 100)(x - 4/3)(x + 15/2))/((x - 3/2)(x - 4/3)(x + 15/2))

5) Plug in x = 0: 100/(-3/2)K = 1, -200/3K = 1, K = -3/200

So one equation that works is

y = -3/200((18x^2 + 28x + 100)(x - 4/3)(x + 15/2))/((x - 3/2)(x - 4/3)(x + 15/2)). You can multiply this out if you want.

Answer 2

When I graph sofarsogood's solution, I get a y intercept of (0,1) rather than (0,-1). If I negate the 3, then the intercept is correct, but the slope of the asymptote is negative instead of positive. Still, I think it is a good effort.

Answer 3

You'll tell your teacher that the term "root" applies only to polynomials which are clearly ruled out here. They are called "zeroes" of a function.

And your function has to be defined on all of the complex plane so that it has non real zeroes? I think the guy is insane.