The function has a horizontal asymptote at y = 3.
The function has a vertical asymptote at x = -3.
The function has a vertical intercept of 1.
The function has a zero of x = -2.
Because the function has a zero at x = -2, and a vertical asymptote at x = -3, therefore
[tex]y= \frac{k(x+2)}{x+3} [/tex]
Write the function in the form
y = [tex]y= \frac{k(1+ \frac{2}{x}) }{1+ \frac{3}{x} } [/tex]
As x → ∞, y → k.
Because a horizontal asymptote exists at y = 3, therefore k = 3.
A vertical intercept exists at y = 1.
The function is
[tex]y = \frac{3(x+2)(x+1)}{(x+3)(x+1)} [/tex]
Answer: [tex]y = \frac{3(x+2)(x+1)}{(x+3)(x+1)} [/tex]
