Question 747339
I'm assuming you have factors of (x-4) and (x+3) on the bottom from the vertical asymptotes.  

For part b, in order to have a horizontal aymptote of y=2, the degree (highest power of the variable) of the numerator and denominator must be the same. When they are, you get a horizontal asymptote that is equal to the ratio of the leading coefficients.  So you just need something with 2x^2 on the top.  

For part c, if the x-intercept is -1, that means the value of the function will be zero at x=-1.  If the function value is going to be zero, the top of the function needs to be zero.  We had 2x^2 on top so far.  If we plug in -1 for x, you get 

{{{2*(-1)^2}}}

{{{2*1}}}

{{{2}}}

Therefore, you'll need to subtract 2 from 2x^2 to get the top to equal zero.

{{{(2x^2-2)/((x-4)(x+3))}}}