## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$A=1$ $B=2$
It is given, that $f(2)=0$. By substituting $x=2$ into the function $f(x)$ we obtain: $f(2)=\frac{2-B}{2-A} \\ 0=\frac{2-B}{2-A}$ $f$ is not defined when the denominator equals $0$. Since $f(1)$ is undefined, the it follows that the denominator is equal to $0$ when $x=1$. Hence, $1-A=0$ $1-A+A=0+A$ $1=A$ Substituting $1$ to $A$ in $0=\frac{2-B}{2-A}$ gives: $0=\frac{2-B}{2-1}\\ 0=\frac{2-B}{1}$ $0(1)=2-B\\ 0=2-B\\ B=2$