PLEASE HELP! Limits and Derivatives, I need help with how to solve this type of question. Let f be the function defined by, f(x)=\frac{ax^2+bx+2}{2x^(2)-8}, where a and b are constants. The graph of f has a horizontal asymptote at y=3 , and f has a removable discontinuity at x=2 .

(a) Show that a=6 and b=−13 .
(b) To make f continuous at x=2 , f(2) should be defined as what value? Justify your answer.
(c) At what value of x does f have a discontinuity due to a vertical asymptote? Give a reason for your answer.