PLEASE HELPVertical asymptotes at x=-3 and x=6, x-intercept at (-2,0) and (1,0), horizontal asymptote at y=-2. Write an equation for the rational function

Answer: y = -2(x+2)(x-1)/((x+3)(x-6)) = (-2x^2 -2x +4)/(x^2 -3x -18)Step-by-step explanation:A polynomial function will have a zero at x=a if it has a factor of (x-a). For the rational function to have zeros at x=-2 and x=1, the numerator factors must include (x+2) and (x-1).For the function to have vertical asymptotes at x=-3 and x=6, the denominator of the rational function must have zeros there. That is, the denominator must have factors (x+3) and (x-6). Then the function with the required zeros and vertical asymtotes must look like ... f(x) = (x+2)(x-1)/((x+3)(x-6))This function will have a horizontal asymptote at x=1 because the numerator and denominator degrees are the same. In order for the horizontal asymptote to be -2, we must multiply this function by -2.The rational function may be ... y = -2(x +2)(x -1)/((x +3)(x -6))If you want the factors multiplied out, this becomes y = (-2x^2 -2x +4)/(x^2 -3x -18)