Nonlinear functions SAT question

Choice C is correct. It's given that $f(x)=\frac{a}{x+b}$ and that the graph shown is a partial graph of $y=f(x)$. Substituting $y$ for $f(x)$ in the equation $f(x)=\frac{a}{x+b}$ yields $y=\frac{a}{x+b}$. The graph passes through the point $(-7,-2)$. Substituting $-7$ for $x$ and $-2$ for $y$ in the equation $y=\frac{a}{x+b}$ yields $-2=\frac{a}{-7+b}$. Multiplying each side of this equation by $-7+b$ yields $-2(-7+b)=a$, or $14-2b=a$. The graph also passes through the point $(-5,-6)$. Substituting $-5$ for $x$ and $-6$ for $y$ in the equation $y=\frac{a}{x+b}$ yields $-6=\frac{a}{-5+b}$. Multiplying each side of this equation by $-5+b$ yields $-6(-5+b)=a$, or $30-6b=a$. Substituting $14-2b$ for $a$ in this equation yields $30-6b=14-2b$. Adding $6b$ to each side of this equation yields $30=14+4b$. Subtracting $14$ from each side of this equation yields $16=4b$. Dividing each side of this equation by $4$ yields $4=b$. Substituting $4$ for $b$ in the equation $14-2b=a$ yields $14-2(4)=a$, or $6=a$. Substituting $6$ for $a$ and $4$ for $b$ in the equation $f(x)=\frac{a}{x+b}$ yields $f(x)=\frac{6}{x+4}$. It's given that $g(x)=f(x+4)$. Substituting $x+4$ for $x$ in the equation $f(x)=\frac{6}{x+4}$ yields $f(x+4)=\frac{6}{x+4+4}$, which is equivalent to $f(x+4)=\frac{6}{x+8}$. It follows that $g(x)=\frac{6}{x+8}$.

Choice A is incorrect. This could define function $g$ if $g(x)=f(x-4)$.

Choice B is incorrect. This could define function $g$ if $g(x)=f(x)$.

Choice D is incorrect. This could define function $g$ if $g(x)=f(x)\cdot (x+4)$.