Question

Question #34779

Answer 1

that means `b=-1`, `c=-12`, and a can be anything except -3 or 4

Explanation:

if the graph has vertical asymptotes at x=-3 and x=4, then the denominator of the function must be `k(x-(-3))(x-(4))=k(x+3)(x-4)` where k is some constant.

since the problem gives the function in the form: `y=(x-a)/(x^2+bx+c)`, the value of k must be 1 so the coefficient of the `x^2` is also 1.

this means the equation is now: `y=(x-a)/((x+3)(x-4))=(x-a)/(x^2-x-12)`

if the numerator is `x+3` or `x-4` then the vertical asymptotes "created" earlier will just become holes, because the `x+3` of `x-4` will factor out from both the numerator and denominator.

that means `b=-1`, `c=-12`, and a can be anything except -3 or 4