Question 269456
Lets go one step at a time.
vertical asymptotes: x = - 1, x = 6
we get a basic idea that our function is
(i) {{{y = (ax+b)/((x+1)(x-6))}}}
-----
horizontal asymptote: y = 3
we know that as x approaches infinity, then the function approaches 3. SInce the denominator is degree 2, the numerator must be degree 2, so we could say
(ii) {{{y = (3x^2+ ax+b)/((x+1)(x-6))}}}
-----
x-intercepts: - 7, 5
If we let x = -7 or x = 5, then the function should become 0.
part 1 - x = -7 gets us
(iii) {{{y = (3(-7)^2 - 7a + b)/(((-7)+1)((-7)-6))}}}
which becomes
(iv) {{{y = (147 - 7a + b)/78}}}
this means that the numerator = 0, so
If a = 6, then b = -105. So our equation is now
(v) {{{y = (3x^2+ 6x - 105)/((x+1)(x-6))}}}
part 2 - x = 5 gets us
(vi) {{{y = (3*5^2+ 5a + b)/((5+1)(5-6))}}}
which becomes
(vii) {{{y = (75+ 5a + b)/(-6)}}}
this means that the numerator = 0, so
If a = 6, then b = -105. The key part is setting the numerator of (iv) = numerator of (vii) to get values for a and b that work
So, our new equation is
(viii) = {{{y = (3x^2+ 6x - 105)/((x+1)(x-6))}}}
-----
hole at x = 0 
this means that we multiply both numerator and denominator by x to get our final answer equation as
(ix) {{{y = ((x)(3x^2+ 6x - 105))/((x)(x+1)(x-6))}}}