Question

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

Answer 1

We have

$f(x)=\frac{x^2-4x+3}{x^2+ax+b}$

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:

$f\mleft(x\mright)=\frac{x^{2}-4x+3}{x^{2}+x-12}$

So the coefficients on the denominator are:

a= 1

b= -12