Question

A= _____B= ______i need the answer for the missing blank

Answer 1

We have

`f(x)=(x^2-4x+3)/(x^2+ax+b)`

We have an asymptote in x=-4 and a discontinuity in x=3. Therefore, the denominator of the function is

`(x+4)(x-3)`

we can expand the expression and obtain the next one

`\begin{gathered} (x+4)(x-3)=x^2-3x+4x-12 \\ =x^2+x-12 \end{gathered}`

The function is:

`f\mleft(x\mright)=(x^(2)-4x+3)/(x^(2)+x-12)`

So the coefficients on the denominator are:

a= 1

b= -12