Answer:
a = 1 and b = -12
Step-by-step explanation:
right from the start, we can observe that there is an vertical asymptote at x =-4, which means that the denominator = 0 at x = -4
at x=-4, denominator becomes
(-4)² + (-4)a + b = 0
4a-b = 16 ------> Eq 1
also notice that there is a removable discontinuity at x =3, this makes sense when we realize that the numerator (x²- 4x+3) can be factorized as (x-3)(x-1)
In order for this removable discontinuity to occur, this means (x-3) must also be a factor of the denominator.
i.e denominator (x²+ax+b) can be factored as (x-3) (x+c) where c is a root of the denominator.
Hence,
(x²+ax+b) = (x-3)(x+c) expand the right side of this equation
x²+ax+b = x² + (c-3)x -3c
Comparing the coefficients, we get
a = c-3 -------> Eq 2
b = -3c ---------> Eq 3
Now we have 3 equations and 3 unknowns, we can solve for a, b and c
If we substitue Eq 2 and Eq 3 into Eq 1,
4a-b = 16
4 (c-3) - (-3c) = 16
4c-12 + 3c = 16
7c = 28
c = 4
substitute this into Eq 2 and Eq 3, we get a = 1 and b = -12
