# Question #34779

Dec 26, 2017

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 $k \left(x - \left(- 3\right)\right) \left(x - \left(4\right)\right) = k \left(x + 3\right) \left(x - 4\right)$ where k is some constant.

since the problem gives the function in the form: $y = \frac{x - a}{{x}^{2} + b x + c}$, the value of k must be 1 so the coefficient of the ${x}^{2}$ is also 1.

this means the equation is now: $y = \frac{x - a}{\left(x + 3\right) \left(x - 4\right)} = \frac{x - a}{{x}^{2} - x - 12}$

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