Well first lets get one with three real zeros.
y = k(x-x1)(x-x2)(x-x3)
but one happens twice
y = k(x-x1)(x-x2)(x-x2)
denominator = 0 at x=-2 and x = 3
y = c[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
when x = 0, y = 1
y =1 = c[-x1 x2^2]/-6
6 = c x1 x2^2
when x and y are big, y = 2x
big x
y = c x^3/x^2 = c x
so c = 2
so
x1 x2^2 = 3 or x1 = x2^2/3
y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
y = 2[(x-x2^2/3)(x-x2)(x-x2)]/[(x+2)(x-3)]
hmmm, unless I am missing something I am still free to choose x2
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.
2 answers
x1 x2^2 = 3 or x1 = 3/x2^2
y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
y = 2[(x - 3/x2^2)(x-x2)(x-x2)]/[(x+2)(x-3)]
y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
y = 2[(x - 3/x2^2)(x-x2)(x-x2)]/[(x+2)(x-3)]