sat-dog

Choice D is correct. The table gives that $f\left(x\right)=0$ when $x=2$. Substituting 0 for $f\left(x\right)$ and $2$ for $x$ into the equation $f\left(x\right)=ax+b$yields $0=2a+b$. Subtracting $2a$ from both sides of this equation yields $b=-2a$. The table gives that $f\left(x\right)=-64$ when $x=1$. Substituting $-2a$ for $b$, $-64$ for $f\left(x\right)$, and $1$ for $x$ into the equation $f\left(x\right)=ax+b$ yields $-64=a\left(1\right)+\left(-2a\right)$. Combining like terms yields $-64=-a$, or $a=64$. Since $b=-2a$, substituting $64$ for $a$ into this equation gives $b=\left(-2\right)\left(64\right)$, which yields $b=-128$. Thus, the value of $a-b$ can be written as $64-(-128)$, which is $192$.

Choice A is incorrect. This is the value of $a+b$, not $a-b$.

Choice B is incorrect. This is the value of $a-2$, not $a-b$.

Choice C is incorrect. This is the value of $2a$, not $a-b$.