Design a rational function with the following characteristics: three real zeros, one of multiplicity 2; y-intercept 1; vertical asymptotes x=-2 and x=3; oblique asymptote y=2x+1.

Answer 1

First, we note that the oblique asymptote of a rational function has the form $y=mx+b$ where $m$ is the coefficient of the highest degree term of the polynomial in the numerator divided by the coefficient of the highest degree term of the polynomial in the denominator. Since the oblique asymptote is $y=2x+1$, this means that the numerator has a degree of 1 higher than the denominator.

Next, we note that the function will have vertical asymptotes at $x=-2$ and $x=3$ since these values cannot be in the domain of the function. This means that the denominator must have factors of $(x+2)$ and $(x-3)$.

Finally, since the function has a y-intercept of 1 and a zero of multiplicity 2 at some value $a$, the numerator must have a factor of $(x-a)^2$.

We can then write the function as follows:

[f(x) = \frac{(x-a)^2}{(x+2)(x-3)}]

To find the value of $a$, we can plug in the y-intercept of 1 and solve for $a$:

[1 = \frac{(a-a)^2}{(1+2)(1-3)} = \frac{a^2}{-4}]

Solving for $a$, we get $a = \pm 2i$.

Thus, the function that satisfies all the given conditions is:

[f(x) = \frac{(x-2i)^2}{(x+2)(x-3)}]