this graph represents the function f(x)=x^2-4x+3/x^2+ax+b.

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

i dont know sorry

step-by-step explanation: