Question

Help please!!! Much appreciated.
Write a rational function with vertical asymptotes at x=-2 and x= 1, a horizontal asymptote at y=2, and a zero at 3.

Answer 1

vertical asymptoes for x must be in the denominator:

f(x) = g(x)/ (x+2)(x-1)
g(x) must contain the same degree as the denominator becasue f has a horizantal asymptotes.
so 3(x-2)/ (x+2)(x-1)

Answer 2

Is that the answer then?

Answer 3

no that one has a zero at 2, you want a zero at 3

Answer 4

denominator should be \((x+2)(x-1)\) as stated above

Answer 5

since that is a polynomial of degree 2, your numerator has to be a polynomial of degree 2 as well, and the leading coefficient has to be 2 to give you the correct horizontal asymptote

Answer 6

first guess would be
\[\frac{2x^2}{(x+2)(x-1)}\] but unfortunately that has a zero at \(x=0\) and you need one at \(x=2\)

Answer 7

try
\[\frac{2(x-2)^2}{(x+2)(x-1)}\]

Answer 8

Then what @satellite73 ?